Proving set of bounded continuous functions is an open set appreciate your help with the below:
Question: 
Let C[0,1] be the set of continuous functions from [0,1] to $\mathbb{R}$. Consider the metric space M = (C[0,1],d) where d denotes the sup metric. Show the set of functions in C[0,1] whose image is contained in (0,1) is an open set in C[0,1].
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My attempt: 
So we let K denote the set of continuous functions from [0,1] to $\mathbb{R}$ whos image is contained in (0,1). Take an arbitrary function k in K. Then since k is bounded, it attains its bounds. So a and b exists in [0,1] such that 0 < k(a) $\leq$ k(x) $\leq$ k(b) < 1 for x in [0,1]. Pick $\epsilon$ = min{ k(a) , 1-k(b) }, so we have the open ball $B_{\epsilon}(h)$ of radius $\epsilon$ around k. 
I am unsure from here onwards. Can you please suggest how should I proceed please? 
Many thanks

In order to show that $K$ is open, you have chosen a point $k\in K$, and need to show that $B_\epsilon(k) \subset K$ for some $\epsilon >0$.  You have chosen $\epsilon = \min(k(x), 1-k(x))$, which guarantees the inequalities $\epsilon \le k(x)$ and $\epsilon \le 1 - k(x)$ for all $x\in [0,1]$.  Note that these inequalities can be rewritten as $0 \le k(x) - \epsilon$ and $k(x) + \epsilon \le 1$.  So far, so good.
In order to complete the proof, you must take an arbitrary $g\in B_{\epsilon}(k)$, and show that $g(x) \in (0,1)$ for all $x\in [0,1]$.  This would show that $g \in K$, which would then imply $B_\epsilon(k) \subset K$, as desired.  I will leave it at that, with a hint to examine the ramifications of the statement '$g\in B_\epsilon(k)$'.
