An incomplete attempt
If $A$ is dense in $B$, we can approximate the members of $B$ arbitrarily well with elements of $A$. In other words, for every $b \in B$ and $\varepsilon > 0$ there exists an $a = a(\varepsilon) \in A$, for which $\Vert b - a(\varepsilon)\Vert < \varepsilon$.
''$\subseteq$''
Now let $x \in \overline A$, meaning either $x \in A$, or $x$ is an accumulation point of $A$. Every element of $A$ is contained in $B$, so we only need to worry about the accumulation points.
Let $x_a$ then be an arbitrary accumulation point of $A$. Therefore there exists a sequence $(x_n) \subset A$, that converges towards $x_a$. But as each $x_n \in A$, and $A$ is dense in $B$, each $x_n$ is arbitrarily close to a certain $b_n \in B$. Then the norm $\Vert b_n - x_n \Vert < \varepsilon$.
Therefore the sequence $(b_n) \to x_a$ as well, making $x_a$ an accumulation point of $B$, as well as $A$. As $B$ is closed, it contains all of its accumulation points, resulting in $\overline A \subseteq B$.
''$\supseteq$''
Let $y\in B$. As $B$ is closed, it contains all of its accumulation points $x_b$, and we can find a sequences $(b_n)$ that converge toward said points. We also know that $A \subseteq B$, so every $a \in A$ is also in $B$.
But then what? I can't immediately think of a way to include $B$ in the closure $\overline A$ with these facts.
$\bar A \supset B$: Consider an arbitrary $y$ in $B$. In the case that $y \in A$, we are done. Otherwise, $A$ is dense in $B$, so for every $\epsilon > 0$, there exists an $x \in A$ with $\|x - y\| < \epsilon$. Thus, for $n = 1,2,3,\dots$, we may select an element $x_n \in A$ such that $\|x_n - y\| < 1/n$.
We see that $(x_n)$ is a sequence converging to $y$, which means that $y$ is an accumulation point of $A$. Thus, $y \in \bar A$.