For this Theorem, the proof on my course notes is like:
Suppose, for a contradiction,that $f: [a,b] \to \mathbf R$ is continuous but not uniformly continuous on $[a,b]$.
Choose $\epsilon \gt 0$ so that for all $\delta \gt 0$ there exisit $x,y \in [a,b]$ such that $|x - y| \le \delta$ but $|f(x) - f(y)| \gt \epsilon$. Thus for each $k \in \mathbb Z^{+}$ choose $x_{k},y_{k}\in[a,b]$ with $|x_{k}-y_{k}| \le \frac{1}{k}$ and $|f(x_{k}) -f(y_{k})|\gt \epsilon$
By the Bolzano Weierstrass Theorem, we can choose a convergent subsequence $(y_{k_{j}})_{k_{j}}$ of $(y_{k})_k$} . Let $c = \lim_{j \to \infty}y_{k_{j}}$. For all $j$ we have $|x_{k_{j}} - y_{k_{j}}| \le \frac{1}{k_{j}}$,hence by sequeeze theorem we have $x_{k_{j}} \to c$.
Since $f$ is continuous at $c$ and $x_{k_{j}} \to c$, $y_{k_{j}} \to c$, we have $f(x_{k_{j}}) \to f(c) ,f(y_{k_{j}}) \to f(c)$ so that $|f(x_{k_{j}}) - f(y_{k_{j}})| \le \epsilon$ giving a contradiction.
I just don't understand why this proof only holds for Closed Interval ? Or where we used some special properties of Closed Interval in the above proof? Can someone help me? Thansk!