Proof that retract of Hausdorff space is closed

[PsychonautQQ]

I'm reading this proof on the matter:
https://math.stackexchange.com/questions/278755/show-that-a-retract-of-a-hausdorff-space-is-closed

How do we know that the final neighborhood they come up with is disjoint from A?

 

[andrewkirk]

I think it is explained in Harald Hanche-Olsen's comment in reply to his own answer. It is the second comment to that answer, and dated Mar 6 '13 at 7:07. Except, I think he should have written $y$ instead of $x$ in it because he is referring to an arbitrary element of U, and $x$ has been fixed.To flesh it out a bit.

1. We chose U and V to be disjoint nbds of x and a, which the Hausdorff property allows us to do.
2. We know $S\equiv r^{-1}(V\cap A)\cap U$ is open because it is the intersection of two open sets. The first set is open because it is the pre-image of a set that is open in A under a continuous map r.
3. Let $y$ be an arbitrary element of S. Then $r(y)\in V$ and $y\in U$ so, since $U,V$ are disjoint, we have $r(y)\neq y$.
4. Hence $y\notin A$ since every element of A maps to itself under $r$.
5. Hence S is an open nbd of $x$ that is disjoint from A.
6. Since $x$ was chosen as an arbitrary point in $X-A$, that means that every element of $X-A$ has an open nbd in $X-A$.
7. Hence $X-A$ is open. Hence $A$ is closed.