Is the closure of the ball of $1$-Lipschitz functions still equi-Lipschitz? $\DeclareMathOperator\Lip{Lip}$Let $\Lip_0(\mathbb R^d)$ be the space of Lipschitz functions $f:\mathbb R^d\to\mathbb R$ vanishing at zero, i.e., $f(0)=0$,  and equipped with the norm $\|f\|:=\|\nabla f\|_{\infty}$. Then $\big(\Lip_0(\mathbb R^d), \|\cdot\|\big)$ is a Banach space. Now we endow $\Lip_0(\mathbb R^d)$ with an alternative topology, denoted by $w$ and generated by the open sets $\mathcal O_{u}(f;\varepsilon)$ as below:
$$\mathcal O_{u}(f;\varepsilon) \quad:=\quad \left\{g\in \Lip_0(\mathbb R^d):~ \left|\int_{\mathbb R^d} \big[\nabla(f-g)(x)\cdot u(x)\big]\right| dx <\varepsilon \right\},$$
where $f\in \Lip_0(\mathbb R^d)$, $u\in L^1(\mathbb R^d;\mathbb R^d)$ and $\varepsilon>0$. Let $\mathcal C:=\{f\in\Lip_0(\mathbb R^d): \|f\|\le 1\}$ and denote by $\overline {\mathcal C}$ its $w$-closure. Could we prove
$$\sup_{f \in \overline {\mathcal C}}~ \|f\| ~<~ \infty?$$ 
 A: $\DeclareMathOperator\Lip{Lip}$The answer is yes, and in fact I claim that $C$ is closed in the $w$-topology. This immediately implies
$$
\sup\limits_{f\in\bar C}\|f\|=\sup\limits_{f \in C}\|f\|\leq 1.
$$

Let me record here a preliminary observation, which will be useful in the sequel.
As per the OP's comment above, the convergence $f_n\to f$ in the $w$-topology simply means that the gradients converge weakly-* in $L^\infty$, in other words
$$
f_n\overset{w}{\to} f
\qquad \mbox{iff}\qquad
\lim\int \nabla f_n\cdot u=\int\nabla f\cdot u
\quad \forall\,u\in L^1.
$$

The proof goes next as follows: Take $f\in \overline C$, meaning that (according to the preliminary observation) there is a sequence $f_n\in C$ such that $\|f_n\|\leq 1$ and $\nabla f_n\overset{*}{\to} \nabla f$.
In particular, observe (by definition of the norm $\|.\|$ on $\Lip_0$) that the sequence $\{\nabla f_n\}$ belongs to the unit ball $B_1^\infty$ in $L^\infty(\mathbb R^d)$, which is weakly-* relatively compact according to the Banach-Alaoglu theorem. By uniqueness of the weak-* limit we conclude that $\nabla f\in B^\infty_1$ too, hence $\|\nabla f\|_\infty\leq 1$ and in fact $f\in C$.
