Extension of Goldstine theorem Is the following claim true?

Claim. Let $E$ be a Banach space and $F$ its closed subspace. Assume $x\in (E\setminus F)\cup\{0\}$ and $y^{**}\in F^{\perp\perp}$, then there exist a net $(y_\alpha)\subset F$ such that $\sup_{\alpha}\Vert x+y_\alpha\Vert\leq\Vert x+y^{**}\Vert$ and $(y_\alpha)$ weak-$^*$ converges to $y$ in $\sigma(Y^{**}, Y^*)$ topology.

To get the Goldstine theorem just take $x=0$.
If claim is untrue what are additional assummptions?
 A: Here is an improved version of counterexample I posted in comments. Consider $c_0$, the space of real sequences $(x_n)_{n=0}^\infty$ that converge to $0$. Under the standard supremum norm, its dual is $\ell_1$ and the second dual is $\ell_\infty$. The  pairing for both dualities is $\langle x,y\rangle = \sum_{n=0}^\infty x_n y_n$. 
Our $E$ will be $c_0$ with  another norm,
$$\|\mathbf x\|_E = \max\left( \frac12 |x_0|, \ \sup_{n\ge 1} |x_n-x_0|\right) \tag{1} $$
It is easy to see that 
$$ \frac12\sup_{n\ge 0}|x_n| \le \|\mathbf x\|_E \le 2\sup_{n\ge 0}|x_n|$$
that is, the norm is equivalent to the original one. Hence, the dualities stay in place except that the dual norms on $E\approx \ell_1$ and $E^{**}\approx \ell_\infty$ will not be the same as $\ell_1$ and $\ell_\infty$ norms. 
I  don't want to compute the  norm of $E^*$, so I will use Goldstine's theorem to show that the new norm on $E^{**} \approx \ell_\infty$ is still given by (1). 
Indeed, Goldstine's theorem implies that for $\mathbf y \in E^{**}$, 
$$\|\mathbf y\|_{E^{**}} = \inf \left(\limsup_k \|\mathbf y_k\|_E\right)$$
where the infimum is taken over all sequences $(\mathbf y_k)$ in $E$ converging to $\mathbf y$ in the weak* topology. (It suffices to look at sequences, because $E^*$ is separable.) By truncating   $\mathbf y$ we get one such sequence, which  shows 
$$\|\mathbf y\|_{E^{**}} \le \max\left( \frac12 |y_0|, \ \sup_{n\ge 1} |y_n-y_0|\right) \tag{2}$$
On the other hand, weak* convergence implies  coordinatewise convergence, and  the right hand side of (2) is lower semicontinuous under coordinatewise convergence (it's a supremum of continuous functions). Therefore, for any sequence $\mathbf y_k$ weakly* converging to $\mathbf y$ we have 
$$
\max\left( \frac12 |y_0|, \ \sup_{n\ge 1} |y_n-y_0|\right) \le \liminf_{k\to\infty} \|\mathbf y_k\|_E
$$
Thus, (2) holds as an equality.
It remains to describe the counterexample: $$\begin{split}F&=\{\mathbf x\in c_0: x_0=0\}\\ F^\perp &=\{\mathbf x\in \ell_1: x_n=0\ \forall n\ge 1\} \\ F^{\perp \perp}&=\{\mathbf x\in \ell_\infty: x_0=0\}\\ \mathbf x&=(1,0,0,0,\dots)\in E\setminus F \\ \mathbf y^{**} &= (0,1,1,1,\dots)\in F^{\perp\perp} \end{split}$$ 
For every $\mathbf y\in F$ we have $\|\mathbf x+\mathbf y\|\ge 1$ because the sequence $\mathbf x+\mathbf y$ begins with $1$ and contains arbitrarily small terms, thus making the supremum part of the norm at least $1$. On the other hand, $\|\mathbf x+\mathbf y^{**}\|=1/2$.


If claim is untrue what are additional assumptions?

I can prove it only in the case when  $F$ is an isometric summand of $E$, that is $E=F\oplus_p G$ for some $p\in [1,\infty]$ and some subspace $G$ of $E$. Of course, this case is pretty much trivial.  
Indeed, $E^{**} = F^{**} \oplus_p G^{**}$, and the subspace $F^{\perp\perp}\subset E^{**}$ is naturally identified with the summand $F^{**}$. 
Given $x$ as in the question, write it as $x_1+x_2$ with $x_1\in F$, $x_2\in G$. Note that 
$$
\|x+y^{**}\| = (\|x_1+y^{**}\|^p + \|x_2\|^p  )^{1/p} 
$$
Applying Goldstine's theorem to $x_1+y^{**}\in F^{**}$ we get a net $y_\alpha $ in $F$ such that 
$\|x_1+y_\alpha\|\le \|x_1+y^{**}\|$ and $y_\alpha \to y$ in weak* topology. Then
$$
\|x+y_\alpha\| = (\|x_1+y_\alpha\|^p + \|x_2\|^p  )^{1/p} \le \|x+y^{**}\|
$$
as required. $\Box$
At first I thought that $F$ being $1$-complemented in $E$ would be enough for the conclusion, but I am not sure about that anymore. 
