Reconstructibility of topological spaces

The answer is ``No''. Let $X$ be the Cantor set and let $Y = X \setminus \{p\}$, where $p$ is any element of $X$. Let $\varphi \colon X \to Y$ be any bijection. Since the Cantor set with one point removed is homeomorphic to the Cantor set with two points removed, the given condition is satisfied. However, $X$ is compact whereas $Y$ is not.

Edit: In response to the question by Loïc Teyssier, there are many ways to see that the assertion is true. One simple way is to note that if two points are removed from the Cantor set, the one-point compactification of the resulting space is compact, metrizable, zero-dimensional, and has no isolated points, and is, therefore, homeomorphic to the Cantor set. Hence, when the point at infinity is removed, the resulting space is homeomorphic to the Cantor set with a single point removed, but it is also the Cantor set with two points removed.

Edit. As pointed out by anonymous, the following argument assumes that the homeomorphisms $X\setminus\{x\}\cong Y\setminus\{\varphi(x)\}$ are all induced by $\varphi$. I will leave it for a while and maybe delete it later.

Recall that a subset $A\subset X\setminus\{x\}$ is open in the subspace topology if and only if there exists an open $V\subset X$ such that $A=V\cap(X\setminus\{x\})$. The subsets $V\subset X$ with $A=V\cap(X\setminus\{x\})$ are $A$ and $A\cup\{x\}$.

Consider $U\subset X$. If $U$ is open in $X$, then $U\setminus\{x\}$ is open in $X\setminus\{x\}$ for all $x\in X$.

On the other hand, let $U\setminus\{x\}$ be open in $X\setminus\{x\}$ for all $x\in X$. Assume that $U$ is not open in $X$, then $U\setminus\{x\}$ must be open in $X$ for all $x\in U$, and $U\cup\{y\}$ must be open in $X$ for all $y\notin U$.

Assume $X$ has at least three distinct elements. Then there are two cases.

If $U$ has at least two elements $x_1\ne x_2$ then $U\setminus\{x_i\}$ must be open in $X$, so $U=(U\setminus\{x_1\})\cup(U\setminus\{x_2\})$ is open, too.

If $X\setminus U$ has at least two elements $y_1\ne y_2$ then $U\cup\{y_1\}$ and $U\cup\{y_2\}$ must be open in $X$, so $U=(U\cup\{y_1\})\cap(U\cup\{y_2\})$ is open, too.