Let $(X, \|\cdot\|_X)$ be a normed vector space and $(X^{\ast},\|\cdot\|_{X^{\ast}})$ its dual space. I have to prove, that
$$ \forall x\in X:\quad \|x\|_X = \sup_{T\in X^{\ast}}\{|T(x)| : \|T\|_{X^{\ast}}=1\}$$
How can I show this using Hahn-Banach theorem?
There's a neat corollary of Hahn-Banach saying that for every $x$ there exists a linear operator $T$ with unitary norm such that $T(x)=||x||$.
Using this and the definition of the norm on the dual space, you have your proof.
I will elaborate on the previous poster's answer by proving the aforementioned corollary, which is often referred to as "$X^*$ norms $X$." In particular:
Claim: For any $x\in X$ there exists $T\in X^*$ such that $\vert\vert T\vert\vert=1$ and $\vert T(x)\vert=\vert\vert x\vert\vert$.
Pf.: Let $L$ be the one dimensional subspace defined by $x$, that is $L=\{\alpha x\mid \alpha\in\mathbb{R}\}$. Let $T_0:L\rightarrow\mathbb{R}$ send $\alpha x\mapsto\alpha\vert\vert x\vert\vert$. It's easy to check that $T_0$ is bounded, linear, and has norm $1$. Thus, we may use Hahn-Banach to extend this functional to $X$, as $T$, while preserving its norm. This $T$ works.
This immediately answers your question because it shows us that $\sup_{T\in X^{\ast}}\{|T(x)| : ||T||_{X^{\ast}}=1\}\geq \vert\vert x\vert\vert$ and it is obvious that $\sup_{T\in X^{\ast}}\{|T(x)| : ||T||_{X^{\ast}}=1\}\leq \vert\vert x\vert\vert$ because if $\vert\vert T\vert\vert=1$ then $\vert T(x)\vert\leq \vert\vert x\vert\vert$.
