How to prove a continuous function is uniformly continuous on a compact set using BW theorem?

Question

Question. Let S be a compact set in $\mathbb R^n$ and $f:S\rightarrow \mathbb R^m$ be a continuous function. Prove that $f$ is uniformly continuous on S.

I want to prove it using Bolzano-Weierstrass theorm.

This is my attempt.

Suppose to the contrary that $f$ is not uniformly continuous.

Then $\Vert x_k-y_k \Vert \rightarrow 0$ and $\Vert f(x_k)-f(y_k) \Vert \nrightarrow 0$ for some $\{x_k\},\{y_k\}\in S$

Since S is compact, $\{x_k\}$ is bounded. Then by Bolzano-Weierstrass theorem, $\{x_k\}$ has a subsequence $\{x_{k_j}\}$ converging to $x\in S$.

Similarly, $\{y_k\}$ has a subsequence $\{y_{k_{j_l}}\}$ converging to $y\in S$.

Also, $\{x_{k_j}\}$ has a subsequence $\{x_{k_{j_l}}\}$ converging to $x\in S$.

Then since f is continuous, $\Vert f(x_{k_{j_l}})-f(x)\Vert \rightarrow 0$ and $\Vert f(y_{k_{j_l}})-f(y)\Vert \rightarrow 0$.

Since $\Vert x_k-y_k \Vert \rightarrow 0$, it follows that $\Vert f(x)-f(y)\Vert \rightarrow 0$. (*)

Then $\Vert f(x_{k_{j_l}})-f(y_{k_{j_l}})\Vert \le \Vert f(x_{k_{j_l}})-f(x)\Vert + \Vert f(x)-f(y)\Vert +\Vert f(y)-f(y_{k_{j_l}})\Vert \rightarrow 0.$ (contradiction)

1. I want to know if my proof is okay.
2. (*) part is just my guess. If it is right, can you please explain why it is justifiable?

Thanks in advance!