Let $X$ be a compact metric space, $(f_n)_{n \in \mathbb{N}} \subset X^X$ a set of homeomorphisms and $f \in X^X$ a homeomrphism. Suppose that $f_n \rightarrow f$ uniformly. Let $E \subset X$ be closed and $U \subset X$ open. Suppose that $E \subset U$. I would like to show that eventually, $f(E) \subset f_n(U)$. Explicitly, I want to show that there exists $N \in \mathbb{N}$ such that for each $n \in \mathbb{N}$, $n \geq N$ implies that $f(E) \subset f_n(U)$.
I have tried a number of ways. I suppose that most promising is by contradiction: Suppose that for each $N \in \mathbb{N}$, there exists $n \in \mathbb{N}$ such that $n \geq N$ and there exists $x \in E$ such that $f(x) \not \in f_n(U)$. Then we obtain a sequence $(x_n)_{n \in \mathbb{N}} \subset E$ such that for each $n \in \mathbb{N}$, $f(x_n) \not \in f_n(U)$.
Since $E$ is closed and $X$ is compact, $E$ is compact. Then there exists a subsequence $(x_{n_k})_{k \in \mathbb{N}}$ and $x \in E$ such that $x_{n_k} \rightarrow x$. By continuity of $f$, $f(x_{n_k}) \rightarrow f(x) \in f(E) \subset f(U)$. So that eventually, $f(x_{n_k}) \in f(U)$. By uniform convergence, \begin{align*} d(f_{n_k}(x_{n_k}), f(x)) &\leq d(f_{n_k}(x_{n_k}), f(x_{n_k})) + d(f(x_{n_k}), f(x)) \\ & \rightarrow 0 \end{align*} So $f_{n_k}(x_{n_k}) \rightarrow f(x) \in f(E) \subset f(U)$, which implies that eventually $(f_{n_k}(x_{n_k}))_{k \in \mathbb{N}} \subset f(U)$
Here I am pretty stuck. I'm not sure if the original claim is true, but if it is, could I get a hint? Im starting to suspect maybe I need stronger conditions like $f_n, f$ being isometries