I am doing my best to understand the proof given to me in my class notes. It is attached below:
Proof
We prove this by contradiction. Suppose that $f$ is continuous on $[a,b]$ but not uniformly continuous. Then there exists $\epsilon > 0$ such that for all $n \in \mathbb{N}$ there exist $x_n, y_n$ such that $|x_n - y_n| < 1/n, |f(x_n) - f(y_n)|> \epsilon$. By the Bolzano-Weierstrass Theorem, there exist $x_{n_k}, y_{n_k}$ which converge to $x, y $ respectively. Hence, using the first inequality: $x=y$. Therefore $f(x) = f(y)$, a contradiction.
Questions:
Firstly, in order for the $x=y$ to work, we need $\lim_{k\to\infty}(x_{n_k} -y_{n_k})= 0 $. Why is this condition true? Does the fact that $x_{n_k}$ and $y_{n_k}$ are subsequences of $x_n$ and $y_n$ imply this?
I took a look at a similar proof on this link: http://www.math.ku.edu/~lerner/m500f09/Uniform%20continuity.pdf which seems to completely discredit the proof in my course notes, as the very first question below is: "Why, in the proof of the theorem, can’t we just take a convergent subsequence of $x_n$ and a convergent subsequence of $y_n$ and proceed directly to the conclusion? "
P.S. I am new to Analysis and a detailed explanation would be appreciated.