$A=\Big\{f \in C[0,1]: f(x)\neq 0,\forall x \in [0,1]\Big\}$ is open in $C[0,1]$ 
Prove: The set $$A=\Big\{f \in C[0,1]: f(x)\neq 0,\forall x \in [0,1]\Big\}$$ is open in $C[0,1]$ with 'sup' metric via open set definition

The author gives the advice " how to choose $r$ with $B(f,r) \subset E$", namely $r=\vert f(0) \vert$ 
From geometrical point of view I know  $B(f,r) \subset E$ but how to prove mathematically?
Ok! Start with , let $g \in B(f,r)$. Then $\text{sup}\;\vert f(x)-g(x) \vert<r$.
How to prove $g(x) \neq 0$ for all $x \in [0,1]$ ?
 A: I disagree with the suggestion of using $r = |f(0)|$.
Instead, first realise that $f(x) > 0$ for all $x$ or $f(x) < 0$ for all $x$, by the intermediate value theorem. Without loss of generality, consider $f(x) > 0$ for all $x$ (by replacing $f$ with $-f$ if necessary).
By the extreme value theorem, $f$ must achieve its absolute minimum somewhere on the interval. In particular, there must be some $x_0 \in [0, 1]$ such that $0 < f(x_0) \le f(x)$ for all $x \in [0, 1]$. I suggest letting $r = f(x_0)$.
Then, if $g \in B(f; r)$, then
$$f(x_0) > \sup_{x \in [0, 1]} |f(x) - g(x)|,$$
and hence using the definitions of $x_0$ and the supremum, for all $x \in [0, 1]$,
$$f(x) > |f(x) - g(x)| \ge f(x) - g(x) \implies g(x) > 0.$$
A: Take $f\in A$, you want to find $\epsilon$ such that $B(f,\epsilon)\subset A$. Since $|f|$ is continuous on the compact $[0,1]$ it has a nonzero minimum $m$, because this minimum will be attained and $f$ is never zero. Then if you take $\epsilon=\frac{m}{2}$ and $g\in B(f,m)$, $|g(x)|=|f(x)-(f(x)-g(x))|\geq |f(x)|-|f(x)-g(x)|\geq m-\frac{m}{2}=\frac{m}{2}> 0$, $\forall x\in [0,1]$. Then $g$ is in $A$ too.
