real analysis - Uniform continuous functions on bounded sets are Lipschitz?

I'm trying to prove the following:

if $f: E \rightarrow \mathbb{R}$ where $E$ is a bounded subset of $\mathbb{R}$, and $f$ is uniformly continuous then there exists $K$ such that
$$|f(x)-f(y)|\leq K|x-y|$$
for all $x,y\in E$

now I have written down this proof which I'm unsure of:

Assume for a contradiction that for all $n \in \mathbb{Z}$ there exists $x_n,y_n \in E$ such that $$|f(x_n)-f(y_n)|\geq n|x_n-y_n|$$

then since $E$ is bounded there exists a convergent sub sequence $\{x_{n_k}\}$ which converges to some number $p$, now since $f$ is uniformly continuous $\{f(x_{n_k})\}$ converges to some number $l$. Thus by the triangle inequality we have that $|f(y_{n_k})-l|\geq ||f(y_{n_k})-f(x_{n_k})|-|f(x_{n_k})-l||\rightarrow \infty$ as $k \rightarrow \infty$ thus $f$ is unbounded which is a contradiction.

Answer

Your main error is in the interpretation of the last formula: $$|f(y_{n_k})-l|\geq |f(y_{n_k})-f(x_{n_k})|-|f(x_{n_k})-l|.$$ We have $|f(y_{n_k})-f(x_{n_k})|\geq n_k|y_{n_k}-x_{n_k}|$ but this does not imply that $|f(y_{n_k})-f(x_{n_k})|\to \infty .$ We have $|y_{n_k}-x_{n_k}|\to 0$ so $n_k|y_{n_k}-x_{n_k}|$ need not go to $\infty.$