Past year paper question.
Let $(X, d)$ be a metric space and let $f: X \to \mathbb{R}$ be a continuous function, where $\mathbb{R}$ is given the standard metric. Assume that for any $\epsilon > 0$, the set $\{ x \in X : |f(x)| \geq \epsilon \}$ is a compact metric subspace of $X$. Show that $f$ is uniformly continuous on $X$.
Attempt:
Let $\epsilon >0$ be given. Let $K := \{ x \in X : |f(x)| \geq \frac{\epsilon}{2} \}$. $f$ is continuous on $K$ $\implies f$ is uniform continuous on $K$. Then there exists $\delta$ such that $d(x,y)<\delta \implies |f(x)-f(y)| < \frac{\epsilon}{2}$.
Take any $2$ points $x,y \in X$, such that $d(x,y)<\delta$. If $x,y \in K$, then $|f(x)-f(y)| < \frac{\epsilon}{2} < \epsilon$. If $x,y \notin K$, then $|f(x)-f(y)| \leq |f(x)|+|f(y)| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} \leq \epsilon$.
But I am stuck here, because what if $x \in K$ and $y \notin K$.
Suppose $f$ is not uniformly continuous. Then there exists $\delta >0$ and sequences $\{x_n\},\{y_n\}$ such that $d(x_n,y_n) \to 0$ but $|f(x_n)-f(y_n)| \geq \delta$ for each $n$. For each $n$ either $|f(x_n)| \geq \delta /2$ or $|f(y_n)| \geq \delta /2$. One of these holds for infinitely many $n$. Suppose $|f(x_n)| \geq \delta/2 $ along a subsequence. The hypothesis tells you that the subsequence lies in a compact set, so it has a convergent subsequence. Along this subsequence both sequences $\{x_n\},\{y_n\}$ converge to the same limit $x$ because $d(x_n,y_n) \to 0$. This contradicts continuity of $f$ at $x$.