Let $(f_n)$ be a sequence of $1$-Lipschitz functions from $(X, d_X)$ to $(Y,d_Y)$ where the first one is compact and the latter is complete (I am not sure if this matters). Let $f_n \to f$ pointwise. What I have already shown is that $f$ is also a $1$-Lipschitz function.
Question: How do I show that $f_n \to f$ uniformly using the coverings definition of compacity.
I know that it is simple to show that for each $\epsilon>0$ there exist a finite number of points $(x_i)$ such that $X$ is covered by $\bigcup_i B(x_i, \epsilon).$
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The way I did it is using the criteria: $f_n \to^{unif} f$ if and only if for each sequence $(x_k)$ of $X$ we have $d_Y(f(x_k), f_k(x_k)) \to_k 0$. If we take into account the fact that $X$ is compact, we can assume that $x_k \to x \in X$. Now $$d(f(x_k),f_k(x_k))\le d(f(x_k),f(x))+d(f(x),f_k(x))+d(f_k(x),f_k(x_k)) \le d(f(x_k),f(x))+d(f(x),f_k(x))+d(x,x_k) \to 0 \text{, Q.E.D.} $$ with the last inequality coming from the fact that $f_i$ are $1$-Lipschitz functions.
Just a suggestion of a proof (without the introduction of other notions):
Suppose $\epsilon>0$. If $f_k\to f$ pointwise, for each $x\in X$, there is an $N(x)$, such that for all $n\geq N(x)$: $d_Y(f(x),f_n(x))<\epsilon$. If $x'\in U_\epsilon(x)$ and $n\geq N(x)$, you have $$d_Y(f(x'), f_n(x'))\leq d_Y(f(x'),f(x))+d_Y(f(x),f_n(x))+d_Y(f_n(x), f_n(x'))<3\epsilon\; ,$$ applying Lipschitz-continuity twice ($f_n$ is 1-Lipschitz, and you have proved that $f$ is 1-Lipschitz.) Now you can proceed with the coverings definition of convergence.