You could not find a proof because it is not true. In fact $BL(X)$ is
separable in the topology of uniform convergence if and only if $X$ is totally bounded.
If $X$ is totally bounded, every Lipschitz function can be extended to its
completion $\tilde{X}$, which is compact. Thus we obtain an isometric
embedding $BL(X) \to C(\tilde{X})$.
If $X$ is not totally bounded, there is an $\epsilon > 0$ and an infinite
$A \subset X$, such that $d(x, y) > \epsilon$ for all distinct $x, y \in A$.
For every $V \subset A$
the function $f_V: x \mapsto \min \{ \epsilon, d(x, V) \}$ is a bounded
Lipschitz function. Clearly $\|f_V - f_W\|_\infty = \epsilon$ for
distinct $V$ and $W$, hence the set of all these functions is discrete.
This shows that $w(BL(X)) \ge 2^{|A|} > \aleph_0$.
Addendum: The above also holds for spaces of uniformly continuous functions, which is the context where it is usually discussed. The fact that a uniformly continuous function on a metric space has a unique
extension to the completion of the space is a special case of the result
discussed in this answer.
This gives us a well-defined mapping $BL(X) \to C(\tilde{X})$ which takes
$f: X \to \mathbb{R}$ to $\tilde{f}: \tilde{X} \to \mathbb{R}$.
That this mapping is an isometry with respect to the sup norm follows easily from the fact that $X$ is dense in $\tilde{X}$ and therefore $f[X]$ is dense
in $\tilde{f}[\tilde{X}]$ by continuity of $\tilde{f}$.
