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Theorem
The uniform limit of continuous functions is continuous. More precisely, let $ (f_n)$ be a sequence of functions on a set $ S \subseteq \mathbf{R}, $ suppose $f_n \longrightarrow f$ uniformly on $S$, and suppose $S = dom(f).$ If each $f_n$ is continuous on $S$, then $f$ is continuous on $S$.

I would like to prove that some $f_n$ is not uniformly continuous. So my question is, if I were to set this up like a proof by contradiction or a proof by contra-positive, would that be sufficient?

i.e. If I wrote: Assume, to get a contradiction, that $f_n \longrightarrow f$ uniformly on $S=dom(f)$ and each $f_n$ is continuous on $S$, then $f$ is not continuous on $S$ .....or something like this?

Is this structure sufficient to contradict "$f_n \longrightarrow f$ uniformly on $S$"

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    You assumed uniform convergence at the beginning, so I think what's being contradicted is the continuity of $f_n$. – 2017-02-26
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    @GNUSupporter okay thanks, that's exactly what I was thinking but really wasn't certain. – 2017-02-26
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    @GNUSupporter What if I state, "assume to get a contradiction that $f_n$ converges to $f$ uniformly and that each $f_n$ is continuous on $S$" ? – 2017-02-26
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    From I see in the edited question body, you may start like this, but the assumption "$f$ is not continuous" will turn out to be redundant: in fact, we can directly prove the continuity of $f$, so it'll no longer be a proof by contradiction, but simply a constructive proof. – 2017-02-26

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