Is this acceptable proof of continuous $f:[a,b]\rightarrow \mathbb R$ being uniform continuity? Pick some $\epsilon \in \mathbb R_+$, now for all $x\in[a,b]$ there exists a $\delta_{x,\epsilon}$, such that $|f(x)-f(x+h)|<\epsilon$, when $h<|\delta_{x,\epsilon}|$.
Define now a sequence $x_n=\inf \{x:\delta_{x,\epsilon}<\frac{1}{n}\}$. This is an increasing sequence by construction and bounded by $b$. We have
$$\sup_n x_n=\lim_{n\rightarrow \infty} x_n=x_\infty \in [a,b].$$
Now $$|f(x_\infty)-f(x_\infty+h)|<\epsilon, \space \space \space 0\leq |h|<\delta_{x_\infty,\epsilon}$$
Hence we have $0<\delta_{x_\infty,\epsilon}$. Conclude by noting that $f$ is continuous so the same applies in the neighborhood, hence for all $x$ we have $\delta_{x_\infty,\epsilon}\leq\delta_{x,\epsilon}$ and hence we can choose $\delta_{\epsilon}=\delta_{x_\infty,\epsilon}$.