Topology: Show restriction of continuous function is continuous, and restriction of a homeomorphism is a homeomorphism

Your proof of Claim 1 is correct, and your proof of Claim 2 is correct but unclear.

First, the last line of your proof ``This shows all homeomorphisms...'' can be omitted. The fact that all homeomorphisms are local homeomorphisms has nothing to do with your claim that $f|_A$ is a homeomorphism.

Second, some of your reasoning in the third paragraph of Claim 2 depends on the fact that the restricted mapping $f|_A$ is a bijection. You could start out by noting that $f|_A$ is a bijection since $f$ is a homeomorphism. Then it suffices to prove that $f|_A$ is continuous and open (since being open in this case is the same thing as the inverse map being continuous).

Now consider the second line in your third paragraph, which starts with $M = A\cap U \ldots$ I don't understand the presence of $Y$ in this line. $f|_A(A) = f(A),$ not $Y$. So you can replace $Y$ with $f(A)$ and then delete the extraneous references to $Y$. Finally, as I alluded to above, your assertion that the image of the intersection of two sets is the intersection of the images is only true because $f|_A$ is a bijection-- this fact isn't true in general. So it might be good to mention that in your proof.

Correct.

It would be better in #2 to say " It suffices to show that $f|_A\to f(A)$ is continuous and open because it is a bijection." First, because, although it is necessary, it matters that the proof is done when it has been shown, because it is sufficient. Second, because you state the properties "continuous" and "open" in the same order that you prove them.

To show that $f|_A\to f(A)$ is open , it is briefer to observe that, because $f$ is a bijection and $A\cap U\subset A,$ we have $f|_A(A\cap U)=f(A\cap U)=f(A)\cap f(U).$