Metrizable space and homeomorphism Show that a topological space $X$ is metrizable if and only if there exists a homeomorphism of $X$ onto a subspace of some metric  space $Y$.
First, assume that there exists an homeomorphism $f$ between a topological space $X$, and a subspace of a metric space $(Y,d)$, define for $a,b$ in $X$
$$d'(a,b)= d(f(a),f(b)) $$
$d'$ is a metric on $X$, How can I prove that  it generates  the topology on $X$?
Also, for the converse, can I just take a metric space $Y$ such that $X\subset Y$ ? or how can I prove that? .
 A: Let $\tau$ be the topology on $X$ and let $A \in \tau$. We have, then, $f(A)$ is open in $(Y,d)$. Let $x \in A$, then $y := f(x) \in f(A)$, so $\exists$ $\epsilon > 0$ such that $B_d(y, \epsilon) \subset f(A)$. We have that: $z \in B_{d'} (x, \epsilon) \iff d'(z, x) < \epsilon \iff d(f(z), y) < \epsilon \iff f(z) \in B_d(y, \epsilon)$. So, $B_{d'}(x,\epsilon) = f^{-1}(B_d(y,\epsilon)) \subset f^{-1}(f(A)) = A$. This shows that $A$ is open in $(X,d')$. Now, just do the backwards process to get the other inclusion.
For the converse, notice that the identity function on $X$ is such a homeomorphism.
A: For your first question see the answer of Ahmed.
The converse can easily cause some confusion. This answer shows a route.
Let $X=\langle S,\tau\rangle$ where $\tau$ denotes the original topology and $S$ denotes the underlying set.
The fact that $X$ is metrizable tells us that there exists a metric $d:S\times S\to[0,\infty)$ such that the topology $\tau_d$ induced by metric $d$ coincides with $\tau$. 
So if $Y:=\langle S,d\rangle$ then the map $S\to S$ prescribed by $s\mapsto s$ is a homeomorphism $X\to Y$. 
$\langle S,d\rangle$ is nothing more than a notation of $\langle S,\tau_d\rangle$ emphasizing that the topology is generated by metric $d$.
Finally observe that metric space $Y$  can be looked at as a subspace of itself.
