Unit ball of continuous functions is a closed set - Proof with neighborhood argument This question is trivial if one uses sequence definition, but I want to use the usual topological definition of closed set. That is , a set is closed if its complement is open.
Let $U=\{f\in C([0,1]):\|f\|_\infty\leq1  \}$ be given. I want to show that $U^c=\{f\in C([0,1]):\|f\|_\infty>1\}$ is open. I can prove by using sequence argument so please help about topological proof.Thanks for any help!

Let $f \in U^c$. Let $\delta = {\|f\|-1}>0$. Then if $g \in \mathrm B_\delta(f)$, then $$\|g\| \ge \|f\| - \|f-g\|>\|f\| - \delta=1$$So $\mathrm B_\delta(f)\subset U^c$.
