This is an exercise from Royden's Real Analysis.
Let $X$ be a normed linear space and $W$ a subspace of $X^*$ that separate points. For any topological space $Z$, show that a mapping $f:Z\to X$ is continuous, where $X$ has the $W$-weak topology, if and only if $g(f):Z\to\mathbb R$ is continuous for all $g \in W$.
I don't know how to use separability of $W$ on $X$ to show that $f$ is continuous.
Basically this is because the elements of $W$ are enough to generate the weak topology.
More in the specific, you want to show that if $g(f)$ is continuous for all $g\in W$ then $f$ is continuous, right?
Let $z_n\to z$. You need to show that $f(z_n)\to f(z)$ for the weak topology.
Suppose $x\in X$ is a weak accumulation point of $f(z_n)$ different from $f(z)$. Thus $L(f(z_{n_k}))\to L(x)$ for all $L\in X^*$. On the other hand $g(f)$ is continuous, for all $g\in W$, so $g(f(z_n))\to g(f(z))$. In particular $g(x)=g(f(z))$ for all $g\in W$.
Since $W$ separates points and we are supposing $x\neq f(z)$, there is $g$ so that $g(x)\neq g(f(z))$, a contradiction. If $f(z_n)$ goes to infinity a similar argument applies.