Does every element of the weak-star closure of a set belong to the weak-star closure of a bounded subset?

The Krein-Smulian theorem says that for a convex set $S$, having weak*-closed intersections with closed balls implies being weak*-closed. The fact that convexity is required here suggests that the answer to your question should be negative. And it is.

Let $X=\ell^2$. In the dual space, also $\ell^2$, consider the "infinite ellipsoid" $$ S = \left\{y : \sum_{n=1}^\infty y_n^2/n^2=1\right\} $$ Every line through the origin meets $S$; therefore, $0$ is in the weak closure of $S$.

On the other hand, $0$ is not in the weak closure of any intersection $S\cap B_R$ where $B_R=\{y:\|y\|\le R\}$. Indeed, consider the weak-open sets $$ U_N = \left\{y: \sum_{n=1}^N y_n^2 < \frac12\right\} $$ If $N$ is large enough, then for every $y\in U_N\cap B_R $ we have $$ \sum_{n=1}^\infty y_n^2/n^2 < \frac12+ \frac{R^2}{(N+1)^2} < 1 $$ hence $U_N\cap (B_R\cap S)= \varnothing$.

In addition to the answer by 404, I have two further counterexamples in $\ell^2$: \begin{align*} A &= \{ \sqrt{n} \, e_n : n \in \mathbb{N}\} \\ B &= \{ e_m + m \, e_n : m,n \in \mathbb{N}, 1 \le m < n \} \end{align*} Again, $0$ is in the weak* closure, but all intersections of $A$ or $B$ with bounded sets are discrete! (It is also interesting to note that this implies that there is no sequence in $A$ or $B$ converging weak*ly to $0$.)