Let $E$ be a normed vector space, $F\subseteq E$ a subspace and $S\in L(F,\ell^\infty)$. Now I have to show that there exists an operator $T\in L(E,\ell^\infty)$ such that $T|_F=S$ and $\|T\| =\|S\|$.
A hint was to look at the functionals $l_n:F\rightarrow\mathbb{K},x\mapsto(Sx)_n$ for $n\in\mathbb{N}$.
So far I have shown that for all $n\in\mathbb{N}$ $l_n\in F'$ (the dual space of $F$). Therefore, by the theorem of Hahn-Banach, there exists $\tilde{l}_n\in E'$ with $\tilde{l}_n|_F=l_n$ and $\|\tilde{l}_n\|=\|l_n\|$.
But now I am stuck and dont know how to construct $T$ from that.
Any hint or advice is appreciated!
EDIT: Could I define $T$ with the $\tilde{l}_n$ as $\tilde{l}_n (x) = (Tx)_n$? Then by the properties of the $\tilde{l}_n$ I have shown that $T$ has to be linear and $T|_F=S$.
After your edit, all that remains to show is that $T$ is bounded and $\|T\| = \|S\|$. We know that $\|l_n\| \leq \|S\|$ for each $n$ and since the Hahn-Banach extension of bounded functionals preserves the operator norm, $\|\tilde{l}_n\| \leq \|S\|$ also.
We then get that $$\|Tx\| = \sup_{n \geq 1} |\tilde{l}_n(x)| \leq \|S\| \cdot \|x\|$$ which implies that $\|T\| \leq \|S\|$ as desired.
It is immediate that $\|T\| \geq \|S\|$ since $T$ extends $S$ and so $\|T\| = \|S\|$.