Consider a bounded continuous function that satisfies $|f(x)|\leq\epsilon$ on a compact set K. I am asked to prove that there is a $\delta$ enlargement of K, $K^\delta$ such that $$|f(x)|\leq2\epsilon ,\forall x\in K^\delta$$ $K^\delta=\{x|d(x,K)<\delta\}$ for some metric.
The hint says uses the compactness of K. I was trying to use the finite covering property of K but the argument does not go through when I try to select a $\delta$. I would like to know if there is some alternative way to prove the above claim.
Hint: for each $x\in K$ there is a $\delta_x>0$ such that $y\in B(x,\delta_x)\Rightarrow f(y)\in B(f(x),\epsilon).$ Then, $K\subseteq \bigcup_{x\in K}B(x,\delta_x)$ and $\mathscr A=\{B(x,\delta_x):x\in K\}$ is a cover of $K$. Reduce to a finite subcover and unravel the definitions.