Let $N,k$ be positive integers, equip $\{1,\dots,N\}$ with the discrete topology, let $K$ be a non-empty compact subset of $\mathbb{R}^k$, and let $(f_n:K\rightarrow \{1,\dots,N\})_n$ be a sequence of continuous functions converging uniformly on $K$ to a continuous function $f:K\rightarrow \{1,\dots,N\}$.
Then, for any non-empty compact set $K'\subseteq \{1,\dots,N\}$, do the pre-images $$ K_n\triangleq f^{-1}_n[K'], $$ converge to $f^{-1}[K']$ in the Hausdorff-Pompieu sense?
Uniform convergence only makes sense if $Y = \{1,\ldots,N\}$ is endowed with a metric. Since $Y$ is finite, all metrics on $Y$ are uniformly equivalent and we may take the standard discrete metric $d(a, b) = 1$ for $a \ne b$ and $d(a,a) = 0$.
The answer is "yes". We have $f_n = f$ for $n \ge n_0$, thus also $K_n = f^{-1}(K')$ for $n \ge n_0$.
To see this, note that $f_n \to f$ uniformly implies that there exists $n_0$ such that for $n \ge n_0$ we have $d(f_n(x),f(x)) < 1$ for all $x \in K$. Thus $f_n(x) = f(x)$ for all $x$, i.e. $f_n = f$ for $n \ge n_0$.