Proving a subset in $C[a,b]$ is open 
I have tried to show that the A' is closed as any sequence in A' must converge in A', but I didn't get anywhere.
Any help would be greatly appreciated!
 A: Each continuous $f : [a,b] \to \mathbb R$ attains a maximum $M$ at some $x_M \in [a,b]$ and a minimum $m$ at some $x_m \in [a,b]$. For $f \in A$ and $x \in [a,b]$ you get $0 < f(x_m) = m \le f(x) \le M = f(x_M) < 1$. Let $\epsilon = \min(m,1-M)$. Now consider $g \in C([a,b])$ such that  $d(g,f) = \sup_{x \in [a,b]} \lvert g(x) - f(x) \rvert < \epsilon$. Then for $x \in [a,b]$
$$g(x) = f(x) + g(x) - f(x) \ge f(x) - \lvert g(x) - f(x) \vert > m - \epsilon > 0$$
$$g(x) = f(x) + g(x) - f(x) \le f(x) + \lvert g(x) - f(x) \vert < M + (1 - M) = 1.$$
A: What you would like to say is that $f(t) < 1$ for all $t$ implies $\sup f < 1$ (rather than $\le 1$). This implication doesn't hold for every set, but it does hold for compact sets. Since $[a,b]$ is compact, the Extreme Value Theorem implies that $f$ achieves its supremum at some point $t^*$ and at that point, $f(t^*) = \sup f < 1$.
So using compactness, we give ourselves some wiggle room around our functions. If $f \in A$ then $0 < \inf f$ and $\sup f < 1$. So if $\varepsilon$ is small enough $\sup f + \varepsilon < 1$ and $0 < \inf f - \varepsilon$. This $\varepsilon$ of wiggle room means that if $\|g - f\| < \varepsilon$ then $g \in A$.
Now you should be able to write these ideas in your own words to show that $A$ is open. If you want to test your understanding: 1) modify these ideas to write a proof that $A^c$ is closed 2) show that $A^c$ doesn't need to be closed if the interval $[a,b]$ is replaced with the non-compact set $(0,1)$ (i.e. find a sequence in $A^c$ that converges to a function in $A$).
A: As I said in my comment under the original post, I think it's easier to prove directly — using the Extreme Value Theorem.
Pick any function $f\in A\subset C[0,1]$. Let's show that there's an open neighborhood of $f$ that still lies in $A$. Since $f$ is continuous on the closed bounded interval $[0,1]$, it attains an absolute maximum and an absolute minimum values on the interval. Let's say $f(u)=M$ is the absolute maximum and $f(v)=m$ is the absolute minimum of $f$ on $[0,1]$, i.e. $u,v\in[0,1]$. By definition of the subset $A$, and since $f\in A$, we know that $0<m\le M<1$. Now let $\varepsilon=\min\{m,1-M\}$. The $\varepsilon$-neighborhood of $f$ still lies in $f$, because all functions that differ from $f$ by less than $\varepsilon$ can't reach neither $0$ not $1$. If you think about it geometrically, it's like a tube of radius $\varepsilon$ around the graph of $f$ on $[0,1]$.
