Why does this uniform continuity follow from continuity?

Question

Let $K$ be a compact subset of $\mathbb{R}$. I'm reading a proof and it says the following:

Let $\epsilon > 0$. Since the mapping $\xi \mapsto e^{\imath \, \xi}$ is continuous, we can pick $\delta>0$ such that $$|1-e^{\imath \, x (\xi-\eta)}| \leq \varepsilon$$ for any $|\xi-\eta| \leq \delta$ and $x \in K.$

Why are we sure such $\delta>0$ exists? Doesn't this need some kind of uniform continuity? Also this holds for all $x$ at once?

Answer 1

$K$ is compact, hence bounded. Then there is some $M > 0$ that $[-M, M] \supseteq K$. By the continuity at $0$, for every $\varepsilon > 0$, there is some $\delta > 0$ such that whenever $|y| < M \delta$, $$ |1 - \mathrm e^{\mathrm i y}| < \varepsilon. $$ Then for any pair $\xi, \eta$ and any $x \in K$, as long as $y = \xi -\eta$ satisfies $|y| = |\xi - \eta| < \delta $, we have $|xy| = |x(\xi - \eta)|< M\delta$, then $$ |1 - \mathrm e^{\mathrm i x(\xi - \eta)}| < \varepsilon. $$

Answer 2

Compact sets are bounded. So there exists $M<\infty$ such that $|x(\xi -\eta)| \leq M |\xi-\eta |$. Now use the fact that $e^{ix}$ is uniformly continuous on compact sets. .