Yes. For each unit vector $v$ in $\mathbb{R}^d$ and for any set $B \subseteq \mathbb{R}^d$, let $$r(B,v) = \inf_{b \in B} \langle b,v\rangle,$$ where $\langle x, y \rangle$ is the inner product of $x$ and $y$. It is not too hard to show that for any $B,C \subseteq \mathbb{R}^d$, $$|r(B,v) - r(C,v)| \leq d_H(B,C),$$ with the appropriate conventions regarding $\infty$.
First I will argue in the case where the $A_n$'s and $A$ are closed convex sets, and then I will extend this to arbitrary convex sets.
For any closed convex set $B$, we have that $$B = \bigcap_{v\text{ unit vector}} \{x \in \mathbb{R}^d : \langle x, v\rangle \geq r(B,v)\}.$$ Let $H(r,v)$ denote the set $\{x \in \mathbb{R}^d : \langle x, v \rangle < r\}$. We clearly have that $B^c = \bigcup_{v\text{ unit vecotr}} H(r(B,v),v)$. It's also not too hard to see that for any fixed unit vector $v$, $d_H(H(r,v),H(r',v)) = |r-r'|$.
Now the crux is this: The Hausdorff metric 'commutes' with arbitrary unions in the following sense, if $\{B_i\}_{i \in I}$ and $\{C_i\}_{i \in I}$ are arbitrary families of sets such that $d_H(B_i,C_i) \leq \varepsilon$ for every $i \in I$, then $d_H\left( \bigcup_{i \in I} B_i, \bigcup_{i \in I} C_i) \right) \leq \varepsilon$.
This means that for each $n$, we have that
$$ d_H(A^c_n ,A^c) = d_H\left(\bigcup_{v\text{ unit}}H(r(A_n,v),v), \bigcup_{v\text{ unit}}H(r(A,v),v) \right) \leq \sup_{v\text{ unit}}d_H(H(r(A_n,v),v),H(r(A,v),v)) = \sup_{v\text{ unit}} |r(A_n,v)-r(A,v)| \leq d_H(A_n,A).$$
So we have that $d_H(A_n^c,A^c)$ goes to $0$ as well.
For arbitrary sequences of convex sets, we just need the following facts:
- For any set $B$, $d_H(B,\overline{B}) = 0$, where $\overline{B}$ is the closure of $B$.
- For any convex set $B \subseteq \mathbb{R}^d$, $d_H(B^c, \overline{B}^c) = 0$.
The first fact is standard. The second is not too hard but requires a little bit of proof.