**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.

very interesting exercisefor an elementary topology course. If $X$ has at least 3 points, the answer is "yes". Since I don't want to spoil the fun, here a hint: To check if some $U\subset X$ is open, necessarily all $U\setminus\{x\}$ must be open in $X\setminus\{x\}$. For the reverse, distinguish two cases: $U$ has at least two elements, or $X\setminus U$ has at least two elements. $\endgroup$ – Sebastian Goette Mar 20 '16 at 14:01no. So if you are sure this is right, I think it would be helpful if you would fill in the details. $\endgroup$ – Nate Eldredge Mar 20 '16 at 19:43twovery interesting exercises in one :D $\endgroup$ – მამუკა ჯიბლაძე Mar 21 '16 at 16:465more comments