Proof of Stone-Wierstrass theorem

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Homework Statement



During the proof of the Stone-Wierstrass theorem we define a particular set and claim it is open. I can't figure out how to show it is open.
I wrote out the proof until that point to show how it is constructed, I apologize for the long windedness.

Assume that X is a compact hausdorf spaceA\subset C(X)is an algebra that sperates points and has a continuosu function in it. then \overline{A}=C(X).

Proof:

Since A separates points then for any a,b\in X there exists h\in A such that h(a)\neq h(b). define g(x)=\frac{h(x)-h(a)}{h(b)-h(a)}. Then g(a)=0and g(b)=1.

for any function f\in A define f_{\left(a,b\right)}(x)=\left(f(b)-f(a)\right)g(x)+f(a)\text{then }f_{\left(a,b\right)}(a)=f(a)\text{ and }f_{\left(a,b\right)}(b)=f(b).

Set \epsilon>0 and define U_{\left(a,b\right)}=\left\{ x\quad|f_{\left(a,b\right)}(x)<f(x)+\epsilon\right\} .

My question is how do i gaurantee that U_{\left(a,b\right)} is open? My professor took it for granted. I'm sure it has to do with the pre-image of f but i can't write it out.

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The Attempt at a Solution


 
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With such a questions, you have to rewrite U_{(a,b)} in such a way so that it is clear that it's open. In this case:

U_{(a,b)}=\{x~\vert~f_{(a,b)}(x)<f(x)+\epsilon\}=\{x~\vert~(f_{(a,b)}-f)(x)<\epsilon\}=(f_{(a,b)}-f)^{-1}(]-\infty,\epsilon[).
 
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