I wanted to show that $X' / U^\perp \cong U'$, for $U$ being a closed subspace of the Banach space $X$.
Therefore I looked at $l: X' / U^\perp \cong U' , x' + U^\perp=[x'] \mapsto x'|_U$.
It is clear that this map is onto, as we just take a $u' \in U'$. Then I can use Hahn-Banach to get an appropriate $x'$ such that $x'|_U = u'$ and in particular $||x'||=||u'||$.
Now I need to show that is isometric: since we always have $\|[x']\|\le \|x'\|$ and by Hahn-Banach we have $\|x'\|=\|x'_{|U}\|$, this shows $\|[x′]\|\le \|x'_{|D}\|$ and it should be always true that $\forall y' \in U^\perp: \|x'-y'\| \ge \|x'_{|D}\|$, hence $\|[x']\|\ge \|x'_{|D}\|$. Is this reasoning correct? The isometric property follows.
It is mostly correct now, but could be written better.
Consider $l:X'/U^\perp\to U'$, which sends each coset of $U^\perp$ to its restriction to $U$. This map is well-defined because all elements of the coset agree on $U$.
The map is surjective, because every functional $f$ on $U$ can be extended to a functional $g$ on $X$ (by Hahn-Banach), and $l([g])=f$. Moreover, we can choose $g$ so that $\|g\| = \|f\|$. Then $\|[g]\|\le \|g\|= \|f\|$. We have shown that $l$ does not decrease the norm. Since the norm of restriction is bounded by the norm of original function, $l$ does not increase the norm either. Thus, $l$ is an isometry.