analysis - Clarification of the proof that if $f$ is continuous on $[a,b]$ then $f$ is uniformly continuous on $[a,b]$

Claim: If $f$ is continuous on $[a,b]$ then $f$ is uniformly continuous on $[a,b]$.

Proof: Suppose that $f$ is not uniformly continuous on $[a,b]$, then there exists $\epsilon > 0$ such that for each $\delta > 0$ there must exist $x,y \in [a,b]$ such that $|x-y| < \delta$ and $|f(x)-f(y)| \geq \epsilon$ .

Thus for each $n \in \Bbb N$ there exists $x_n,y_n$ such that $|x_n-y_n| < \frac {1} {n}$, and By B-W, $x_n$ has a subsequence $x_{n_k}$ with limit $x_o$ that belongs to $[a,b]$.

Now here is where I am confused, the author next states that "Clearly we also have $x_0$ is the limit of the sub-sequence $y_{n_k}$". Are $x_n$ and $y_n$ not potentially different sequences? Why would the same indexes chosen to form a sub-sequence give the same convergent value.

Answer

For every $c>0$, there exists $k_0$ such that $k>k_0$ implies that $|x_0-x_{n_k}|k_0$such that${1\over n_{k'_0}}k'_0$,$|x_0-y_{n_k}|\leq |x_0-x_{n_k}|+|x_{n_k}-y_{n_k}|\leq |x_0-x_{n_k}|+{1\over n_k}