Proof: weakly convergent sequence is bounded In class we had to proof:
Let $X$ be normed vector space.
If $(x_n)_n$ is a weakly convergent sequence in $X$ it follows that $(x_n)_n$ is bounded.
While I unterstand the proof we did in class (use of isometric canonical embedding and Hahn-Banach) I want to also check an other approach with a followup Separation Corollary of Hahn-Banach:
Let $X$ be a non-trivial normed vector space.
Then for any $x ∈ X$ there is a functional $x^∗ ∈ X^∗$ with $||x^∗|| = 1$ and
with $x^∗(x) = ||x||$.
Proof for which I need validation:
If $(x_n)_n$ converges weakly to $x$ in $X$, then for any $n$ we have that $x_n - x$ is also in $X$.
Therefore, by the separation Corollary there exists a functional $x^*$ in $X^*$ such that $x^*(x_n-x)=||x_n-x||$.
If we take norm on both sides we get
$||x_n-x||=||x^*(x_n-x)||$
And using linearity of $x^*$:
$||x_n-x||=||x^*(x_n-x)||=||x^*(x_n)-x^*(x)||$
Since $(x_n)_n$ converges weakly to $x$ we have that for any $\epsilon > 0$ there exists an $N$ such that for all $N\leq n$ we have
$||x^*(x_n)-x^*(x)||<\epsilon$
and hence also
$||x_n-x||=||x^*(x_n)-x^*(x)||<\epsilon$
Therefore, $x_n$ converges to $x$ and hence $x_n$ must be bounded.
 A: $(1).$ The canonical embedding of $X$ into $X^{**}$ with $x\to x^{**}\in X^{**},$ where $x^{**}(y)=y(x)$ for all $y\in X^*$ and $x\in X$, is an isometry. That is, $\|x\|=\|x^{**}\|.$
Proof: By the Hahn-Banach Theorem, for  $x\in X$  the functional $y$ on the vector subspace generated  by $\{x\}$, such that $y(x)=\|x\|$, extends to some $y_x\in X^*$ with $\|y_x\|=1.$
Now $\|x^{**}\|=\sup \{|x^{**}(y)|:\|y\|=1\}=\sup \{|y(x)|:\|y\|=1\}\le \|x\|.$
But $\|x^{**}\|\ge |x^{**}(y_x)|=|y_x(x)|=\|x\|.$
$(2).$ The Uniform Boundedness Principle(UBP). If $B$ is a Banach  space and $F\subset B^*$ such that $\sup_{f\in F}\,|f(b)|<\infty$ for each $b\in B$ then $\sup_{f\in F}\,\|f\|<\infty.$
$(3).$ If $(x_n)_{n\in \Bbb N}$ is a weakly convergent sequence in $X$ then $\lim_{n\to \infty}y(x_n)$ exists for each $y\in X^*.$ A convergent series of scalars (i.e. members of $\Bbb R$ or $\Bbb C$) is bounded.
So $\sup \{|x^{**}_n(y)|:n\in \Bbb N\} =\sup \{|y(x_n)|:n\in \Bbb N\}<\infty$ for each $y\in X^*.$ 
Applying the UBP with $B=X^*$ and $F=\{x_n^{**}: n\in \Bbb N|\}$ we have $$\infty>\sup \{\|x^{**}_n\|:n\in \Bbb N\}=\sup \{\|x_n\|:n\in \Bbb N \}.$$
