The answer is pretty close to yes. As Joriki points out, there's some issues involving closure, but what we can say is this:
If $A_n \to A$ and $B_n \to B$ under the Hausdorff (pseudo)metric on $2^X \setminus \{\emptyset\}$ and $A_n \subseteq B_n$ for all $n$, then $\overline{A} \subseteq \overline{B}$.
Suppose that $\overline{A} \not\subseteq \overline{B}$. Then there exists some $x \in \overline{A} \setminus \overline{B}$. Since $x \in X \setminus \overline{B}$, which is open, there exists an open ball around $x$ exists that is disjoint from $\overline{B}$ and hence $B$. Moreover, in this ball, there must exist some point in $A$. Without loss of generality, by replacing $x$ with this new point, possibly reducing the radius, let us suppose that $x \in A$, but $r > 0$ is such that $B(x; r) \cap B = \emptyset$.
Now, since $A_n \to A$, we know that
$$\sup_{a \in A} d(a, A_n) \to 0$$
as $n \to \infty$. So, some $M$ exists such that
$$n \ge M \implies \sup_{a \in A} d(a, A_n) < \frac{r}{3} \implies d(x, A_n) < \frac{r}{3}.$$
We can then fix $a_n \in A_n$ such that $d(x, a_n) < \frac{r}{3}$.
We also know, since $B_n \to B$, that
$$\sup_{b \in B_n} d(b, B) \to 0$$
as $n \to \infty$. In particular, there must exist some $N$ such that
$$n \ge N \implies \sup_{b \in B_n} d(b, B) < \frac{r}{3}.$$
As $B_n \subseteq A_n$, we then have
$$n \ge N \implies \sup_{b \in A_n} d(b, B) < \frac{r}{3} \implies d(a_n, B) < \frac{r}{3}.$$
Thus, if we fix any $n \ge N$, we have
$$d(x, B) \le d(x, a_n) + d(a_n, B) < \frac{r}{3} + \frac{r}{3} < r.$$
But this implies there exists some $b \in B$ such that $d(x, b) < r$, which would place $b \in B(x; r) \cap B = \emptyset$. We have a contradiction, and the result is proven.