If I understand the claims of the OP correctly, I don't think that such a section can actually exist (if there is a misunderstanding on my part, I will happily retract this answer!).

Upon reading the question, I immediately thought of topological vector space versions of the Michael continuous selection theorem (for instance, Theorem 1.2 of Michael's paper *Three mapping theorems*, Proc. Amer. Math. Soc. **15** (1964), 410-415.) and Mori Zippin's later work on operator extension problems, which uses continuous selections.

In one of Zippin's papers in particular, namely *The embedding of Banach spaces into spaces with structure*, Illinois J. Math. **34(3)** (1990), 586-606, we see that the existence of a $w^\ast$-continuous selection $E_1^\ast\longrightarrow F^\ast$ is equivalent to $E$ being *almost complemented* in $F$; by *almost complemented* we mean that there exists a number $\lambda>0$ such that for every compact, Hausdorff $K$ and operator $T: E\longrightarrow C(K)$, there is a continuous linear extention $\overline{T}: F\longrightarrow C(K)$ of $T$ such that $\Vert \overline{T}\Vert \leq \lambda\Vert T\Vert$.

So the opposition to the claim of the OP comes from the existence of Banach spaces admitting subspaces that are not almost complemented. From the definition of almost complemntedness, the most easily illustrated examples are found by taking compact Hausdorff spaces $K$ and $L$ such that $C(L)$ contains an uncomplemented subspace isomorphic to $C(K)$, and considering the operator $T$ to be the identity operator on the the uncomplemented copy of $C(K)$. One example of such spaces $K$ and $L$ is obtained by taking $K=\gamma\mathbb{N}$ (one-point compactification) and $L=\beta \mathbb{N}$ (Stone-Cech compactification). Another example is to take $K=L = \omega^\omega+1$ (the set of ordinals not exceeding $\omega^\omega$, equipped with its natural compact, Hausdorff order topology).

Whilst writing my answer I noticed that Andreas Thom posted an answer mentioning the Bartle-Graves theorem, which also crossed my mind when thinking of the answer to this question. Conincidentally, only weeks ago I was wondering whether Bartle-Graves can be done $w^\ast$-continuously for the adjoint of an isometric inclusion map, and now I can see that the answer is *no*... so I, for one, am very grateful for this question!