TITLE: Continuous choice of Hahn-Banach extensions
QUESTION [6 upvotes]: QUESTION RETRACTED - My original argument was fundamentally mistaken (mixing up lower and upper semi-continuity).  Sorry (and thanks for the useful comments)
I need, and (unless I am seriously mistaken) can prove, the following:
Let $E \subseteq F$ be an (isometric) inclusion of Banach spaces, and let $E^*_1$, $F^*_1$ denote the closed unit balls of their respective duals.  Then the restriction map $F^*_1 \to E^*_1$ admits a weak$^\*$-continuous section from $E^*_1$ to $F^*_1$.
On the unit sphere this section gives you norm-preserving extensions.
This must be known - can anyone provide a reference? (am not a Banach space theorist)

REPLY [3 votes]: After Phil's answer and the ensuing discussion, the remaining question can be formulated as:
Estimate $\lambda$ s.t. if $F\subset E$ and $E/F$ is finite dimensional, then every norm one linear operator $T$ from $F$ into a $C(K)$ space can be extended to an operator from $E$ into the $C(K)$ space which has norm at most $\lambda$.
I want to point out that $\lambda$ cannot be less than two.  I think that it is known that $\lambda$ can be anything larger than two, but I did not find a reference after a quick search.  I called Morry Zippin's attention to the problem; maybe he will comment. Or maybe this was proved after he worked on almost complementation; perhaps by Jesus Castillo and colleagues.  I more or less thought through that $\lambda$ can be anything larger than three. 
Before saying why $\lambda$ cannot be less than two, let's further reformulate the problem.  The second dual of every $C(K)$ is $1$-injective, so the problem reduces to the case where $F$ is a $C(K)$ space and $T$ is the identity operator on $F$. That is, we need to estimate how well complemented a $C(K)$ must be in a superspace in which the $C(K)$ space has finite codimension. In fact, the superspace of $C(K)$ can be assumed to be in $C(K)^{**}$, but that is not particularly useful in getting the lower bound on $\lambda$. 
Consider the $C(K)$ space $c$, the space of convergent sequences under the sup norm. (I'll treat the real case, but the complex case is only slightly more complicated.)  The superspace $E$ is the span in  $\ell_\infty$ of $c$ and the sequence $x$ defined by $x(n)= (-1)^n$.  Let $P_n$ be the projection onto the span of the first $n$ unit vectors in $\ell_\infty$ and let $P$ be any projection from $E$ onto $c$.  Write $Px = a\cdot 1 + y$ with $y$ in $c_0$ and where $1$ is the constant one function. Now for every $n$, the norm of $x-2P_n x$ is one, and $P(x- 2P_n x)=a 1 + y - 2P_n x$.  Since $y$ is in $c_0$, letting $n\to \infty$ gives $\|P\| \ge |a|+2$.  The choice $a=0$, $y=0$ of course gives a projection of norm two.
EDIT Nov. 29, 2011: Zippin's article in Handbook of the Geometry of Banach Spaces, vol. 2, contains information about almost complementation (differently named) including many open problems.