VC dimension of complement
First, observe that it is enough to prove that $d\le d'$. Then, the converse inequality follows from the fact that $\overline{\overline{C}}=C$, and by applying the initial argument to $\overline{C}$.

To prove the claim, we actually prove something stronger: we prove that if $C$ shatters $S$, then $\overline{C}$ also shatters $S$.

Let $S$ be a set that is shattered by $C$, and consider $\Pi_{\overline{C}}(S)$. Let $T\subseteq S$, then there exists $c\in C$ such that $c\cap S=S\setminus T$ (since every subset of $S$ can be reached this way). Let $c'=X\setminus c\in \overline{C}$, then $c'\cup c=X$. Thus, $c'\cap S=T$. Therefore, $\Pi_{C}(S)\subseteq\Pi_{\overline{C}}(S)$.
Applying the same argument to $\overline{C}$, we get $\Pi_{\overline{C}}(S)\subseteq \Pi_{\overline{\overline{C}}}(S)=\Pi_{C}(S)$, and thus we have the equality we wanted, and we conclude the claim.