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I have stubled upon the following proof on Wikipedia. I wonder if it is right.enter image description here

First I think that S must be open interval, right? That is not explicit. Second what troubles me is the a* definition I think it should be a* interval should be open on c so that $\upmu-\epsilon$ would be higher than $f(a*) -\epsilon$, since \upmu never reaches $f(b).

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    Yes, $S$ will be an open interval. – 2017-02-02
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    The fact that $S$ is open follows from the assumptions of the intermediate value theorem: $f(x)$ is continuous on $[a,b]$. This is because $\{x: \ f(x)$f^{-1}((-\infty,u))$ and $(-\infty,u)$ is an open set. – 2017-02-02
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    The statement in the proof is not technically wrong, as you suggest in fact such $a^{*}$ must be in $(c-\delta,c)$, but then clearly $a^{*}\in (c-\delta,c]$ – 2017-02-02
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    @HanulJeon : You're wrong. The set $S$ will in some cases be a union of open intervals. – 2017-02-02
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    @MichaelHardy I should use the term "open set". – 2017-02-02
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    $S$ does not have to be an open set. It could well be neither open nor closed. Nor is it particularly relevant. – 2017-02-02
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    Can you clarify, what is \upmu, you use it twice. – 2017-02-02
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    For any $\epsilon >0$ there exits $a^*,a^{**}$ such that $u-\epsilon – 2017-02-02
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    Sorry for my math terminology but how do you write mu? Thank everyone for the enlightening comments. – 2017-02-02

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The set $S$ is not necessarily an interval, but it is open, which is what is needed for the proof; this follows from the continuity of $f$.

It is true that $a^*$ can never be equal to $c$ (because of $S$ being open). But the interval doesn't play any role. All that you need is that $c-\delta