Show that every closed linear subspace $ \ Y \ $ of a Banach space $ \ X \ $ is weakly sequentially closed Show that every closed linear subspace  $ \ Y \ $ of a Banach space $ \ X \ $ is weakly sequentially closed ,
that is ,  $ \ Y \ $ contains weak limits of all weakly convergent sequences $ \ \{y_n \} \ $ in $ \ Y \ $. 
Answer:
We know closed subspace of a Banach space is also a Banach space.
Thus $ \ Y \ $ is a Banach space or Completely normed linear space.
That means every cauchy sequence in $ \ Y \ $ converges in it.
Now let $ \ \  y_n \to \ y \ $ weakly.
To finish the proof , we have to show that the weak limit $ \ y \ \in Y $ .
But I got no way to show it.
Help me doing this.
 A: $Y$ is weakly closed (strongly closed linear subspaces are weakly closed, this follows from Hahn-Banach, essentially; see the argument here e.g.).
In any topology (so also the weak topology on a Banach space), being closed implies being sequentially closed:
if $A$ is closed and $a_n \to a$, then let $O$ be an open neighbourhood of $a$.
By convergence, $O$ contains a tail of $(a_n)$, so in particular, $O \cap A \neq \emptyset$. As $O$ was arbitary, $a \in \overline{A} = A$, so $A$ is sequentially closed.
So $Y$ is weakly sequentially closed.
A: Let $Y$ be a closed subspace of $X$. Then $Y$ is weakly closed (and hence weakly sequentially closed).
To see this, consider its annihilator $Y^0 = \{f  \in X^* : f|_Y = 0\} \subseteq X^*$. We have
$$Y = \bigcap_{f \in Y^0} \ker f$$
Namely, clearly $Y \subseteq \ker f$ for every $f \in Y^0$ so $Y \subseteq \bigcap_{f \in Y^0} \ker f$.
Conversely, let $x \in X \setminus Y$. A corollary of Hahn-Banach states that there exists $f \in X^*$ such that $f(x) = 1$ and $f|_Y = 0$. Then $x \notin \ker f$ so we conclude $\bigcap_{f \in Y^0} \ker f \subseteq Y$.
Now, for $f \in X^*$ notice that $\ker f$ is a weakly closed set as a preimage of $\{0\}$ by a weakly continuous function $f$. Hence $Y$ is weakly closed as an intersection of weakly closed sets.
