Existence of a path between two sides of a square, which is contained in the intersection of two open subsets of it

Yes. Indeed, shrinking sets $A$ and $B$, if necessary, we can assure that $A\cup B$ covers $I^2$, $A\subseteq [0,1)\times I$, and $B\subseteq (0,1]\times I$. Pick a number $0<\delta<1/2$ such that $3\delta$ is smaller than Lebesgue’s number of the cover of the square bu $A$ and $B$. Triangulate the square as shown in the picture, assuring that each side of each small triangle is at most $\delta$.

Mark triangle vertices belonging to horizontal sides of the square by $0$’s and belonging to vertical sides of the square by $1$’s (the corners of the square mark arbitrarily by $0$’s or $1$’s). Mark an inner triangle vertex $v$ by $0$ if all triangles incident to $v$ are covered by the set $A\cap B$, and by $1$, otherwise. Now, starting from the left bottom corner of the square, begin to walk traversing small triangles, entering and leaving them through edges marked by distinct numbers. It is easy to check that we never traverse any triangle twice. Also it is easy to check that we can stop because we came to a blind alley, only at a corner of the square. It follows that our walk connects either the bottom and the top sides (1) of the square of its left and right sides (2). Consider these cases.

(1) This case is shown at the picture. It is easy to show that the walk has a subwalk $S$ which begins in a triangle adjacent to the bottom side of the square, ends in a triangle adjacent to the top side of the square, and has all intermediate traversed triangles adjacent neither to the bottom nor to the top sides of the square. Each time when $S$ traverses a triangle side, this side has a vertex marked $0$, which follows that both triangles incident to the side are contained in $A\cap B$. Thus $S$ provides a path from the bottom to the top side of the square, contained in $A\cap B$.

(2) It is easy to show that the walk has a subwalk $S$ which begins in a triangle adjacent to the left side of the square, ends in a traingle adjacent to the right side of the square, and has all intermediate traversed triangles adjacent neither to the left nor to the right sides of the square. Let $s_1,\dots, s_n$ be the sequence of triangle sides traversed by $S$. For each $i=1,\dots, n$, a side $s_i$ has a vertex $v_i$ marked $1$. It means that there exists a point $p_i\in I^2\setminus (A\cap B)$ contained in a triangle incident to $v_i$. Let $d$ be the usual distance on $I^2\subseteq\Bbb R^2$. Then $d(v_i,p_i)\le\delta$. Since $d(v_i,v_{i+1})\le\delta$ for each $i=1,\dots, n-1$, by the triangle inequality we have that $d(p_i,p_{i+1})\le 3\delta$ for each $i=1,\dots, n-1$. Pick a point $p_0$ contained in the intersection of the beginning triangle of $S$ and the left side of the square and a point $p_{n+1}$, contained in the intersection of the ending triangle of $S$ and the right side of the square. Then $d(p_0, p_1)\le 2\delta$, $d(p_n, p_{n+1})\le 2\delta$, $p_0\in A$, $p_{n+1}\in B$. So there exists an index $0\le i\le n$ such that $p_i\in A$ (and so $p_i\not\in B$) and $p_{i+1}\in B$ (and so $p_{i+1}\not\in A$). But since $d(p_i, p_{i+1})\le 3\delta$ is smaller than Lebesgue’s number of the cover of the square by $A$ and $B$, a set $\{p_{i}, p_{i+1}\}$ must be contained either in $A$ or in $B$, a contradiction.

PS. The picture and the idea of the proof were taken from Problem 8 of [Sh], on the draw impossibility in the hex game.

References

[Sh] Shashkin Yu. A., Fixed points, Moscow, Nauka, 1989, in Russian.

Since $(\{0\} \times I) \cup B^c$ and $(\{1\} \times I) \cup A^c$ are closed and disjoint, they can be precisely separated by a continuous function $f: I^2 \to I$. Let $C$ be the connected component of $(I \times \{0\}) \cup f^{-1}([\frac13, \frac23])$ that contains $I \times \{0\}$. Being a connected component of a closed set, $C$ is closed. If we suppose that $C$ and $I \times \{1\}$ are disjoint, they too can be precisely separated by a continuous function $g: I^2 \to I$.

The map $(f, g): I^2 \to I^2$ sends each edge of the square to the same edge, so it sends a path $p$ that winds once around the square’s edges to another such path. Continuously contracting $p$ to a point also contracts $(f, g)(p)$ to a point, and such a contraction must sweep through every point in the square. So $(f, g)$ is surjective.

Choose a sequence of points $x_n \in I^2$ such that $(f, g)(x_n) = (\frac12, \frac{1}{n})$. By Bolzano–Weierstrass, some subsequence of $x_n$ converges to a point $x \in I^2$. Then $(f, g)(x) = (\frac12, 0)$, so $x \in C$, and in fact a connected neighborhood of $x$ in the open set $f^{-1}((\frac13, \frac23))$ lies in $C$. But since $g(x_n) \ne 0$, the sequence $x_n$ can never reach that neighborhood, which contradicts its convergence to $x$.

Therefore, $C$ and $I \times \{1\}$ are not disjoint, which means there’s some connected component of $f^{-1}([\frac13, \frac23])$ that intersects both $I \times \{0\}$ and $I \times \{1\}$. Enlarging it to a connected component of the open set $f^{-1}(A \cap B)$ gives an open component, which is therefore path-connected.

The cleanest proof of this I know is to use a Mayer-Vietoris sequence. You can do directly on a square using relative homology, but it's a bit cleaner to note that the original statement is equivalent to asking that for any pair of open subsets $A$ and $B$ of the cylinder $S^1\times I$ such that $A$ contains $S^1\times \{0\}$ and $B$ contains $S^1\times \{1\}$, there exists some loop $A\cap B$ which is not contractible in $S^1\times I$. You can get this equivalence by gluing the top and the bottom of the square together and adding in a strip along this glue line which is in both $A$ and $B$. There's some little details to take care of, but it's possible to ensure that the subspace of the cylinder outside of the extra strip in both $A$ and $B$ is just your original square with its original subsets. Once you have your non-trivial loop, you can find that it must, at some point, traverse over the part of the cylinder that is your original square - which gives you your answer.

That taken care of, we can just write out one part of the Mayer-Vietoris sequence for $A$ and $B$: $$H_1(A\cap B)\overset{f}{\rightarrow} H_1(A)\oplus H_1(B) \overset{g}{\rightarrow} H_1(S^1\times I)$$ where the first map $f$ is the sum of the two embeddings $A\cap B\rightarrow A$ and $A\cap B\rightarrow B$ and where the second map $g$ takes $h_1\oplus h_2$ to the difference of the embeddings of $h_1$ and $h_2$ into $H_1(S^1\times I)$.

The image of this first map must equal the kernel of the second. However, if we let $\alpha\in H_1(A)$ be the loop around $S^1\times \{0\}$ and $\beta\in H_1(B)$ be the similarly oriented loop around $S^1\times \{1\}$, it's clear that these two loops embed to the same thing in $H_1(S^1\times I)$, hence $\alpha\oplus \beta \in \ker g$. As a result, there is some class $h\in H_1(A\cap B)$ such that the embedding of $h$ into $H_1(A)$ gives $\alpha$. Clearly, any chain in the homology class of $h$ contains some loop which is non-trivial in $S^1\times I$, so we are done.