If $X$ is separable, then $\mathcal{F}(X)$ is also separable, where $\mathcal{F}(X) = \overline{span \{ \delta_x : x \in X \}}$ Suppose $X$ is a Banach space. For any $x \in X$, define the set $\mathcal{F}(X) = \overline{span \{ \delta_x : x \in X \}}$ where $\delta_x(f)=f(x)$ for all $f \in$ Lip$_0(X)$. 
The set Lip$_0(X)$ is the set of all real-valued Lipschitz functions which vanish at $0$.
Note that $\delta_x$ is an evaluation functional on Lip$_0(X)$.
I am wondering whether the following statement is true or not.

If $X$ is separable, then $\mathcal{F}(X)$ is also separable.

The statement above is taken from here, so I believe the statement is true. But I have no idea on how to prove it. 
UPDATE: The following is my attempt:
Take the set $\{ \delta_x : x \in X \}$ with the norm $\| \delta_x \| = \| x \|$. Note that $\{ \delta_x : x \in X \}$ is isometric to $X$. Hence, $\mathcal{F}(X)$ is separable. 
Is it correct?
 A: Suppose $X$ is over $\mathbb{R}$.
Let $A$ denote a countable dense subset of $X$. Let $y$ be an element of  ${\cal F}(X)$, and let $\varepsilon>0$. By definition, there exists a finite linear combination $\sum_{k=1}^n a_k \delta_{x_k}$ that belongs to an open ball of radius $\varepsilon/2$ around $y$ in $\mathrm{Lip}_0(X)^*$. For every $f$, and every choice of $z_k\in A$:
$$\left|\left(\sum_{k=1}^n a_k(\delta_{x_k}-\delta_{z_k})\right)(f)\right|=\left|\sum_{k=1}^n a_k(f(x_k)-f(z_k))\right|\leq||f||_{Lip}\sum_{k=1}^n|a_k|\,\|x_k-z_k\|_X$$
Hence,
$$\left\|\sum_{k=1}^n a_k(\delta_{x_k}-\delta_{z_k})\right\|_{\mathrm{Lip}_0(X)^*} \leq \sum_{k=1}^n|a_k| \, \|x_k-z_k\|_X$$
As $A$ is dense in $X$, we can choose $z_k$ so that the r.h.s is less that $\varepsilon/2$. It follows that the distance between $\sum_{k=1}^n a_k\delta_{z_k}$ and $y$ in $\mathrm{Lip}_0(X)^*$ is at most $\varepsilon$. One further approximation argument shows that the set of finite linear combinations $\sum_{k=1}^nr_k\delta_{z_k}$ with rational $r_k$'s, is dense in ${\cal F}(X)$. As this set is countable, it follows that ${\cal F}(X)$ is separable.
