A metric that makes $l^\infty$ separable

with something like $$ d(x,y)=\sum_n2^{-n}|x(n)-y(n)|, $$ the collection of bounded sequences is separable, for instance the rational span of the "standard basis" $e_k(n)=\delta_{kn}$ is dense.

as noted in the comments, the topology generated by this metric is that of pointwise convergence (i.e. bounded sequences as a subspace of $\mathbb{R}^{\mathbb{N}}$ with the product topology).

This answer concerns the more general problem.

Claim: A non-empty set $X$ can be endowed with a separable and metrizable topology if and only if $\# X\leq\mathfrak \#\mathbb R$.

Proof: My comment above explains necessity. As for sufficiency, suppose that $\# X\leq\mathfrak \#\mathbb R$ and let $f:X\to \mathbb R$ be an injective function. Let $Y\equiv f(X)$. Consider the topology $\tau$ on $X$ generated by sets of the form $\{f^{-1}(U)\,|\,U\subseteq Y,\text{ $U$ open in the relative topology}\}$. Then, $f$ becomes a homeomorphism between $X$ and $Y$, where $Y$ is endowed with the relative topology induced by the standard Euclidean topology on $\mathbb R$. It is clear that this topology on $Y$ is $T_1$, regular, and second countable and thus so is the topology $\tau$ on $X$. Now invoke Urysohn's metrization theorem to conclude that $(X,\tau)$ is a separable and metrizable topological space. $\blacksquare$

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In particular, $\#\ell^{\infty}=\#\mathbb R$, so...