Finding a separable, closed subspace. I'm dealing with the following problem from Fabian's book:

Let $X,Y$ be Banach spaces and $\mathrm T\in\mathcal B(X,Y)$. Show that if $Y$ is separable and $\mathrm T$ is onto $Y$, then there is a separable closed subspace $Z$ of $X$ such that $\mathrm T(Z)=Y$.

I figured out that I need to point out a dense, countable subset of $Y$, but after that I'm stuck. Any ideas?
 A: Le $M$ be the kernel of $T$. Then $X/M$ is isomorphic and homeomorphic  to $Y$ (via the map $x+M \to Tx$). Since $Y$ is separable it follows that $X/M$ is separable. Let $(x_n+M)$ be a countable dense subset. Let $Z$ be the closed subspace of $X$ spanned by $(x_n)$. I leave it to you to verify that this $Z$ satisfies the requirements. 
A: As $Y$ is separable, so there exists a countable dense subset $D$ of $Y$; thus $D$ is a countable subset of $Y$ such that the closure of $D$ in $Y$ equals $Y$ itself.
Let's put
$$ D^\prime \colon= T^{-1}(D). $$
Then as $T$ is onto, so we obtain 
$$ T\left( D^\prime \right) = T\left( T^{-1}(D) \right) = D, $$
from which it follows that $D^\prime$ must be countable.
Now let $Z^\prime$ equal the intersection of all the subspaces of $X$ containing $D^\prime$. then $Z^\prime$ is the span of $D^\prime$ (i.e. the set of all possible linear combinations of the elements of $D^\prime$). 
Let $Z$ denote the closure in $X$ of $Z^\prime$. Then $Z$ is a closed subspace of $X$ and $Z \supset D^\prime$.
As $T$ is onto, so we have 
$$ T(Z) \supset  T(D^\prime) = T\left( T^{-1}(D) \right) = D, $$
and therefore
$$ Y \supset \overline{T(Z)} \supset \overline{D} = Y. $$
Thus we have 
$$ \overline{T(Z)}  = Y. \tag{1} $$
But as $T$ is continuous and as $Z$ is closed, so we obtain
$$ T(Z) \subset \overline{T(Z)} \subset T\left( \overline{Z} \right) = T(Z). $$
Thus we have
$$ \overline{T(Z)} = T(Z). \tag{2}$$
From (1) and (2) we obtain
$$ T(Z) = Y. $$
I'm sorry but I can't right now think of how to show that $Z$ is actually separable.
