Let $X$ be a real Banach space and $Y\subset X$ be a (closed) subspace of $X$. Assume that a sequence $y\_n^\*\in S\_{Y^\*}$ weak\*-converges to some $y^\*\in S\_{Y\*}$. (Here $S\_{Y^\*}$ stands for the dual unit sphere.)


Let $x\_0 \in X \setminus Y$ and define $Z:= \textrm{span} (Y\cup \{x\_0\})$.


Is the following true?


There exist a sequence $z\_n^\*\in S\_{Z^\*}$ with $z\_n^\*\restriction\_Y = y\_n^\*$, and $z^\*\in S\_{Z^\*}$ with $z^\*\restriction\_Y = y^\*$, such that $z\_n^\*$ converges weak\* to $z^\*$.


Remarks:


1. It is easy to obtain that a SUBSEQUENCE of $z\_n^\*$ converges weak\* to $z^\*$.
2. I'd like to consider general Banach spaces without additional assumptions (e.g. reflexivity, separability).
3. I wouldn't mind using an additional set-theoretic axiom consistent with ZFC.



Extending a weak*-converging sequence onto a superspace
The norm preserving condition, is as Mikhail points out, essential to prove that the such an extension does not exist. The question is stated for Banach spaces, but notice that the same question for normed spaces is equivalent to the question for Banach spaces because a bounded sequence of functionals converges weak$^\*$ if it converges pointwise on a dense set. 


So assume that the question has an affirmative answer in the category of normed spaces. We claim that a stronger thing is true; namely, in the set up of the OP, the superspace $Z$ of $Y$ can be taken to be $X$ itself. Indeed, the union of a nested family of extensions that are weak$^\*$ convergent gives extension to the union of the domains of the extending functionals, so one can Zornicate just as in the standard proof of the Hahn-Banach theorem. More explicitly, consider all $(Z, (z\_n^\*)\_n)$ such that $Y \subset Z $, $Z$ is a subspace of $X$, and $(z\_n^\*)\_n$ are linear functionals in the unit ball of $Z^\*$ such that for all $n$, $z\_n^\*$ extends $y\_n^\*$, and $(z\_n^\*)\_n$ converges pointwise on $Z$. Partially order by $(Z, (z\_n^\*)\_n) \le (W, (w\_n^\*)\_n)$ provided $Z\subset W$ and for all $n$, $z\_n^\* \subset w\_n^\*$, and observe that the hypothesis in Zorn's Lemma applies. 


To see that the OP's question has a negative answer, let $X=\ell\_\infty$, $Y=c$ (the space of convergent sequences under the supremum norm), and let $y\_n^\*$ be the unit vector basis for $\ell\_1 \subset c^\*$. There are not uniformly bounded weak$^\*$ convergent extensions to $\ell\_\infty$ because weak$^\*$ converging sequences in $\ell\_\infty^\*$ are weakly convergent.


EDIT 12/30/17: If you want an explicit example in the separable setting to the OP's original formulation, again let $X=\ell\_\infty$, $Y=c$ (the space of convergent sequences under the supremum norm), let $y\_n^\*$ be the unit vector basis for $\ell\_1 \subset c^\*$, and let $z\_0 $ be the element of $\ell\_\infty$ whose odd coordinates are all $1$ and whose even coordinates are all $-1$.


