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Let $S$ be a topological space, prove that if $f: S → \mathbb{R}$ is continuous at $a$ and a sequence $x_n∈ S → a$, then $f(x_n) → f(a)$.

We need to prove that, for any $ε > 0$, there exists an $N$ such that $|f(x_n)$ - $f(a)$| < ε $\forall$ $n$ > $N$.

Let $ε > 0$ be given. Choose $δ > 0$ so that if $|x - a| < δ$, then $|f(x) - f(a)| < ε$. So for every $x$ inside this interval $f(x)$ will be inside our desired interval from $f(a)$.

Choose $N_2$ so that $|x_n - a| < δ$ for all $n > N_2$. We can do so because $x_n → a$. Then $|f(x_n) - f(a)| < ε$ $ \forall n> N_2$.

And we have found an $N$ that satisfies our constraint.

Is this proof rigorous enough? Thanks.

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    You seem to know about dollar signs, but you're not using them enough. Enclose _all_ mathematics in dollar signs to make it look better. For instance, `$\forall n>N, |f(x_n) - f(x)| < \epsilon$` gives $\forall n >N, |f(x_n) - f(x)| < \epsilon$. – 2017-02-07
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    Yes, your proof looks good. You could clarify what $S$ is - presumably some subset of $\mathbb R$ or $\mathbb C$ where $|.|$ makes sense ? – 2017-02-07

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Your proof is logically correct, but working with $|x-a|<\delta$ you have tacitly assumed that $S$ is a metric space.

In the formulation of the problem $S$ is just a topological space. Hence you can talk about neighborhoods of points $a\in S$, open sets, etc., but there is no means to talk about a numerical size of such neighborhoods.

We have to prove the following: Given any $\epsilon>0$ there is an $n_0\in{\mathbb N}$ such that $\bigl|f(x_n)-f(a)\bigr|<\epsilon$ for all $n>n_0$, without talking about distances in $S$.

This proof goes as follows (the logic is the same as yours): Since $f$ is continuous at $a$ there is a neighborhood $U$ of the point $a$ such that $\bigl|f(x)-f(a)\bigr|<\epsilon$ for all $x\in U$. Since $\lim_{n\to\infty} x_n=a$ there is an $n_0$ such that $x_n\in U$ for all $n>n_0$. It follows that $\bigl|f(x_n)-f(a)\bigr|<\epsilon$ for all $n>n_0$.