Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!
Let $f:X \to Y$ be continuous and $X,Y$ be metric spaces where $X$ is compact. Then $f$ is uniformly continuous.
My attempt:
Let $d_1, d_2$ be metrics on $X,Y$ respectively.
Assume the contrary that $f$ is not uniformly continuous. Then there is $\delta >0$ such that $$A_n =\{(x,y) \in X^2 \mid d_{1}(x,y) < 1/n \, \text{and} \, d_{2}(f(x), f(y)) \ge \delta\} \neq \emptyset, \quad \forall n \in \mathbb N^*$$ By Axiom of Countable Choice, there is a sequence $(x_n,y_n)_{n \in \mathbb N^*}$ such that $d(x_n,y_n) < 1/n$ and $d(f(x_n), f(y_n)) \ge \delta$ for all $n \ge 1$. Because $X$ is compact, there are subsequences $(x_{n_k})_{k \in \mathbb N^*}$ and $(y_{n_k})_{k \in \mathbb N^*}$ such that $x_{n_k} \to a$ and $y_{n_k} \to b$ as $k \to \infty$.
We have $d_1$ is continuous, so $\lim_k d_{1}(x_{n_k}, y_{n_k}) = d_{1}(\lim_k x_{n_k},\lim_k y_{n_k}) = d_{1}(a,b)$. On the other hand, $\lim_k d_{1}(x_{n_k}, y_{n_k}) \le \lim_k (1/ n_k) = 0$. Hence $a-b=0$ or $a=b$.
Because $f$ and $d_{2}$ are continuous, $\lim_k d_{2}(f(x_{n_k}), f(y_{n_k})) = d_{2}(f(\lim_k x_{n_k}), f(\lim_k y_{n_k})) = d_{2}(f(a), f(b)) = d_2(f(a), f(a)) =0$. On the other hand, $d_{2}(f(x_{n_k}), f(y_{n_k})) \ge \delta$ for all $k \in \mathbb N$, so $\lim_k d_{2}(f(x_{n_k}), f(y_{n_k})) \ge \delta >0$. This is a contradiction.
