I start by considering the complement $$F^c = \{f\in C([0,1]): f(s) < 0, \textrm{for some } s\in[0,1]\}$$ and I want to show that for every $f\in F^c$ the open ball $B(f,r)\subset F^c$ for some positive $r>0$ i.e the set $F^c$ is open in $\left(C([0,1]), d_{\infty}\right)$, where $d_{\infty}$ denotes the max-metric $d_{\infty}(f,g) =\max_{t\in[0,1]}|f(t)-g(t)|$
I pick $g\in B(f,r)$ so by definition we have that $d_{\infty}(f,g) < r$. This means also that $$|f(s)-g(s)| \le \max_{t\in[0,1]} |f(t)-g(t)| < r$$
where $s\in[0,1]$ is fixed s.t $f(s)<0$.Because $$g(s)-f(s) \le |g(s)-f(s)| = |f(s)-g(s)| < r$$
we can for example choose $r = \frac{f(s)}{2}$ which means that
$$g(s) < f(s) + r < \frac{3}{2}f(s) < 0$$
for some $s\in[0,1]$. So we have that $B(f,r)\subset F^c$ and therefore $F^c$ must be closed.
I'm rather uncertain about my proof in general and would like to receive suggestions if it isn't correct