Is my proof for this claim acceptable? (if f is continuous, than f is sequentially continuous) 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. 

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 $n>n_0$. 
