Proof of "removing a point" strategy I am doing a self-study of Hatcher and my topology is failing me.  Specifically at the bottom of p35 he uses a "removing a point" strategy for proving two shapes are not homeomorphic.  (Similarly, this strategy is used in Math StackExchange article 24873, and elsewhwere.)
Intuitively, the strategy seems valid enough, but I am looking for a proof.
Formally, I'd like to prove:
$f:A\rightarrow B$ is a homeomorphism between $A$ and $B$ iff $A-\{x\}$ is homeomorphic to $B-\{f(x)\}$, for every $x$ in $A$.
 A: The $(\impliedby)$ direction is false. Consider the space $X = \{a,b\}$ with the discrete topology (i.e. all subsets are open), and the space $Y = \{c,d\}$ with the topology $\{\varnothing, \{c\}, Y\}$ (so the singleton $\{c\}$ is open but not closed). Then $f : X \to Y$, $f(a) = c$ and $f(b) = d$, is clearly not a homeomorphism. However, it satisfies the condition of your question: the restrictions $\{a\} \cong \{c\}$ and $\{b\} \cong \{d\}$ are both homeomorphisms.
The other direction $(\implies)$ is true whether the two spaces are Hausdorff or not, though, as the proof of Alex S shows.
So logically, you can use the $(\implies)$ direction to show that if you have two spaces $X$ and $Y$, and if for some $x \in X$ and for all $y \in Y$, $X \setminus x$ and $Y \setminus y$ are not homeomorphic, then $X$ and $Y$ are not homeomorphic. Note that it's not sufficient to show that for some $x$ and some $y$ the spaces $X \setminus x$ and $Y \setminus y$ are not homeomorphic; $[0,1]$ is clearly homeomorphic to itself, but $[0,1] \setminus 0$ is not homeomorphic to $[0,1] \setminus \frac{1}{2}$. But even if $X \setminus x$ is homeomorphic to $Y \setminus y$ for all $x$ and $y$, it's possible for $X$ and $Y$ not to be homeomorphic at all.
A: I'm going to edit the statement of your theorem a little bit: $f:A\to B$ is a homeomorphism if and only if the restriction $f|_{A\setminus\{x\}}$ is a homeomorphism for all $x$. I think this is a reasonable restatement. Unfortunately, I cannot prove it unless we add the condition that $A$ and $B$ be Hausdorff. I think this is still helpful, though, as I have always seen the remove-a-point argument applied to subsets of $\mathbb R^n$, CW-complexes, and simplicial complexes. 
$(\Rightarrow)$ Suppose that $f:A\to B$ is a homeomorphism. Let $x\in A$, and let $y=f(x)$. Then the restriction of $f$ to $A$ is still bijective. Additionally, it is well known that the restriction of a continuous function to a smaller domain is still continuous. Thus, $f|_{A\setminus\{x\}}:A\setminus\{x\}\to B\setminus\{y\}$ is still continuous, and similarly for $f^{-1}$. Thus, $f|_{A\setminus\{x\}}$ is still a homeomorphism. Thus, $A\setminus\{x\}$ is homeomorphic to $B\setminus \{y\}.$
$(\Leftarrow)$ Now suppose that $f:A\to B$ and $f_x=f|_{A\setminus\{x\}}:A\setminus\{x\}\to B\setminus \{f(x)\}$ is a homeomorphism for all $x$. Suppose $f(x)=f(y)$. Then $f_x(y)$ does not lie in its codomain unless $x=y$ (making $f_x(y)$ an ill defined evaluation). Thus, $f$ is injective. Now choose $y\in B$ and some $x\in A$. If $f(x)=A$, no worries. If $f(x)\neq y$, then consider $f_x$. It is surjective, so there exists some $x'\in A$ so that $f_x(x')=f(x')=y$. Thus, $f$ is surjective.
Now we must show $f$ is continuous. Suppose that $U\subset B$ is open. If $U=B$, then $f^{-1}(U)=A$, which is open. If $U\neq B$, then there exists $y\in B$ such that $y\not\in U$. Let $f(x)=y$. Then $f_x$ is a homeomorphism, so $f_x^{-1}(U)=V$ is open in $A\setminus \{x\}$. Since $y\not\in U$, $f_x^{-1}(U)=f^{-1}(U)=V$. Since $V$ is open in $A\setminus \{x\}$, there exists an open set $W\subset A$ such that $W\setminus\{x\}=V$. Since $A$ is Hausdorff, $V$ is open in $A$, so $f$ is continuous. A symmetric argument shows that $f^{-1}$ is also continuous, so $f$ is a homeomorphism.
So there is a proof in the Hausdorff case. I wonder if this statement might be false in the non-Hausdorff case. I will try to think of a counter-example.
