Show that there doesn't exists a space $X$ such that $X\times X=S^{2}$.

Here is an algebraic topology free answer, which uses several other big hammers instead.

Assume for a contradiction that $f:X\times X\rightarrow S^2$ is a homeomorhpism.

Consider the involution $\sigma:X\times X\rightarrow X\times X$ which switches the factors: $\sigma(x_1,x_2) = (x_2,x_1).$ Then $\rho = f\circ \sigma \circ f^{-1}$ is an involution of $S^2$. According to Eilenberg (in a proof which doesn't use algebraic topology), $\rho$ is conjugate to a smooth map $\rho '$ in $Homeo(S^2)$ (Thanks to Balarka for noticing and fixing a previous gap here). By averaging a Riemannian metric on $S^2$ we may assume that $\rho'$ is an isometry on $S^2$. In particular, the fixed point set of $\rho'$ must be an embedded submanifold in $S^2$. Note that the fixed point set of $\rho'$ is homeomorphic to the fixed point set of $\rho$.

Now, the fixed point set of $\sigma$ is the diagonal $\Delta X = \{(x, x)\in X\times X: x\in X\}$ so is clearly homeomorphic to $X$. Since $f$ is a homeomorphism, we see the fixed point set of $\rho$, which is equal to $f(X)$, is homeomorphic to $X$. Thus, we have shown that $X$ itself has the structure of a smooth manifold.

From the classification of compact 1-manifolds (which can be done via Riemannian geometry using the exponential map), it follows that $X$ is homeomorphic to $S^1$. Therefore, $S^2$ is homeomorphic to $S^1\times S^1$.

Now, how do we show this is a contradiction without algebraic topology? First, equip $S^2$ and $S^1\times S^1$ with their standard differentiable structures.

In this paper of Hatcher, he shows that any homeomorphism between smooth surfaces is isotopic to a diffeomorphism without using any results from algebraic topology. Thus, we may assume $S^2$ is diffeomorphic to $S^1\times S^1$.

Finally, we note that $S^1\times S^1$ is parallelizable (obvious) while $S^2$ is not (less obvious), so they can't be diffeomorphic. And while the standard proof of the Hairy Ball Theorem uses algebraic topology, Milnor has provided a proof which uses only some analysis and a bit of point set topology.

Assume it were true. Let $h : X \times X \to S^2$ be a homeomorphism. $X$ must have more than one point, so choose two distinct points $a, b \in X$. $X' = h(X \times \lbrace a \rbrace)$ is a homeomorphic copy of $X$ and a retract of $S^2$ (since $X \times \lbrace a \rbrace$ is a retract of $X \times X$). Since $s = h(a,b) \notin X'$, $X'$ is also a retract of $S^2 \backslash \lbrace s \rbrace$ which is contractible. Therefore $X$ is contractible, hence also $X \times X \approx S^2$. This is a contradiction.