Understanding the proof of if $f$ is continuous on a compact set $K$ then $f$ is uniformly continuous on $K$

Question

I am trying to understand the proof of the theorem that if $f$ is continuous on a compact set $K \subseteq \mathbb{R}$ then $f$ is uniformly continuous on $K$. Here is the proof:

I am stuck on a couple things:

(1) How is $\lim [(y_{n_k})-(x_{n_k})] = 0$?

(2) The last statement of the proof claims that this proof has produced the desired contradiction. However, I don't understand how $\left| f(x_n) - f(y_n) \right| \geq \epsilon_0$ was contradicted by concluding that $\lim_{k \to \infty} \left| f(x_{n_k}) - f(y_{n_k}) \right| = 0$.

(3) [Edited from (2)] How does $\lim_{k \to \infty} \left| f(x_{n_k}) - f(y_{n_k}) \right| = 0$ imply $ \left| f(x_{n_k}) - f(y_{n_k}) \right| \geq \epsilon_0$ (in other words, where did the $\lim_{k \to \infty}$ part disappear)?

Any help is greatly appreciated!

Answer 1

First question: If a sequence tends to $0$ so does every subsequence. Since $y_n-x_n \to 0$ so does $y_{n_k}-x_{n_k}$.

Second question: If $|c_n| \geq \epsilon_0$ for all $n$ then $|c_{n_k}| \geq \epsilon_0$ for all $k$. This implies that $c_{n_k}$ cannot tend to $0$.

Answer 2

According to the theorem as $ \lim_{n\to\infty} |x_n -y_n| \to 0$ we have $ |f(x_n)-f(y_n)| \to 0 $ the proof above considers the first part to be true and by contradiction shows that the second part has to be true in case of a function defined on a compact set. As both $x_n$ and $y_n$ converge to $x$ we must have by the Algebraic Limit Theorem that $f(x_n)%$ and $f(y_n)$ converges to $f(x)$. Thus there exist an $n > N_0$ such that $|f(x_n)-f(y_n)|<\epsilon$ , hence the contradiction.