What is the point of this? If $f:X\to Y $ is an homeomorphism then $\forall A\subset X, f|_A:A\to f(A)$ is an homeomorphism What is the point of saying this 
If $f:X\to Y $ is an homeomorphism then $\forall A\subset X, f|_A:A\to f(A)$ is an homeomorphism ?
I don't see the need to make that observation since it's pretty obvious to notice it, if  something works on the whole space $X$ then it's going to work too in a smaller set of $X$. Or am I missing something? Am I right in my reasoning?
 A: You need to work on the subspace topologies: $\tau_{A}=\{A\cap G: G~\text{is open in }X\}$, $\tau_{f(A)}=\{f(A)\cap H: H~\text{is open in }Y\}$. Personally I don't think that is completely obvious, different topologies can turn continuity into discontinuity and open map to non-open map.
A: To see that the restriction of a homeomorphism $\varphi : X\to Y$ to a set $A$ is a homeomorphism, we (at least naively) need to show three things:


*

*$\varphi|_{A}$ is bijective,

*$\varphi|_{A}$ is continuous, and

*$(\varphi|_{A})^{-1}$ is continuous.


Typically, the first result should follow from more general arguments that have already been put forth.  However, for the sake of completeness, note that $\varphi|_{A}$ is surjective on $\varphi(A)$ by construction, and if $\varphi|_{A}(x) = \varphi|_{A}(y)$, then $\varphi(x) = \varphi(y)$, which implies that $x=y$ by the injectivity of $\varphi$.  Hence $\varphi|_{A}$ is injective.  This proves (1).
For (2), suppose that $V \subseteq \varphi(A)$ is open.  Then there is some open $V' \subseteq Y$ such that $V = V'\cap \varphi(A)$.  Since $\varphi$ is continuous, it follows that $\varphi^{-1}(V')$ is open.  On the other hand,
$$(\varphi|_{A})^{-1}(V) = \varphi^{-1}(V') \cap A. $$
But $\varphi$ is continuous, and so $\varphi^{-1}(V')$ is open.  This implies that $(\varphi|_{A})^{-1}(V)$ is open in the subspace topology on $A$, which further implies that $\varphi|_{A}$ is continuous, which completes the proof of (2).
The proof of (3) is, mutatis mutandis, the same as (2).  Indeed, we might summarize both (2) and (3) by the lemma

Lemma:  The restriction of a continuous map $\varphi: X \to Y$ to a subspace $A \subseteq X$ is continuous.

A: We can simply use the definition of the subspace topology on $A:$ it is the weakest one for which the insertion $i:A\to X$ is continuous. Then, $f|_A=f\circ i$. Now, if $V$ is open in $Y$, $(f\circ i)^{-1}(V)=i^{-1}\circ f^{-1}(V).$ Since $f^{-1}(V)$ is open in $X,\  i^{-1}( f^{-1}(V))$ is open in $A$ by definition of the subspace topology and so $f|_A$ is continuous.  The exact same reasoning shows that $(f|_A)^{-1}:f(A)\to A$ is continuous. Since $f|_A$ is bijective with inverse $(f|_A)^{-1}$, we conclude that $f|A$ is a homeomorphism onto $f(A).$
