Let $C_b(\mathbb{R})$ stand for the set of all continuous and bounded self-maps on $\mathbb{R}$, and view this as a metric subspace of $B(\mathbb{R})$, the set of all functions on real numbers. Is $\{f \in C_b(\mathbb{R}): f>0\}$ an open set in $C_b(\mathbb{R})$?
The discussion here indicates that if $\mathbb{R}$ is replaced by some compact set, it would be true. However, real numbers are not compact as $(n-1, n+1)$ for all integers $n$ could cover them.
I am not sure if I found a proper counterexample.
How about $f: (0, 1] \rightarrow (0,1]$ such that $f(x)=x$?
Then for some $\delta>0$ define $g=x-\delta$. It would be within the epsilon ball of $f$, but $g(x)<0$ for some $x$. Is this a counterexample? I feel something is wrong here.