My work:
If we assume $A\bot B\mid C$ and $A\bot B\mid C^c$ hold, then
\begin{align} P(A\cap B\mid C) &= {P(A \cap B \cap C) \over P(C)}\\ &= {P(A\cap C) \over P(C)}{P(B\cap C) \over P(C)}\\ &\Rightarrow P(A \cap C) = {P(A \cap B \cap C)P(C)\over P(B \cap C)} \end{align}
and
\begin{align} &P(A \cap C^c) = {P(A \cap B \cap C^c)P(C^c)\over P(B \cap C^c)}\\ \Rightarrow & P(A \cap C) = P(A) - {(P(A \cap B)-P(A \cap B \cap C))(1 - P(C))\over P(B) - P(B \cap C)}\\ \end{align}
By examining the second expression for $P(A \cap C)$ we see that the $(1-P(C))$ term forces $P(A \cap C) < P(A)$, whereas no such condition was necessary with $A\bot B\mid C$ alone.
Is my thought process correct?
You simply need to show the existence of some probability space where $\small\mathsf P(A\cap B\mid C)=\mathsf P(A\mid C)\mathsf P(B\mid C)$ but $\small \mathsf P(A\cap B\mid C^\complement)\neq\mathsf P(A\mid C^\complement)\mathsf P(B\mid C^\complement)$.
Make one. Take the probability space of $\Omega=\{1,2,3,4,5,6,7\}$ and $\mathsf P\{\omega\}=(1/7)\mathbf 1_{\omega\in\Omega}$.
Find events $A, B, C$ in this space which satisfy both expressions above.