Proof that continuous function respects sequential continuity Let $A\subset\mathbb{R}^n$, $f:A\to \mathbb{R}^m$ be a continuous function. I want to prove that if a sequence $(x_k)\subset A$ converges to some $a\in \mathbb{R}^n$ then $f(x_k)$ converges to $f(a)$. Here's my attempt:
Since $f$ is continuous, $\forall \varepsilon>0,\exists \delta >0 $  such that $\|f(x)-f(a)\|<\varepsilon$ whenever $\|x-a\|<\delta$. Also, since $(x_k)$ converges to $a$, $\forall\varepsilon' > 0$, $\exists k_0 \in\mathbb{N}$ such that $\|x_k-a\|<\varepsilon'$ whenever $k\ge k_0$. Let $\varepsilon'>0$ and set $\varepsilon'' = \varepsilon+\varepsilon'$, and fix $k\ge k_0$. Since $f$ is continuous, then $\exists \delta' >0$ such that $\|f(x)-f(x_k)\|<\varepsilon'$ whenever $\|x-x_k\|<\delta'$. Now, $\|f(x_k)-f(a)\|=\|f(x_k)-f(x)+f(x)-f(a)\|\le$
$\|f(x)-f(x_k)\|+\|f(x)-f(a)\|<\varepsilon'+\varepsilon= \varepsilon''$. Hence, $f$ respects sequential continuity.
Do you think my proof is OK? Thanks.
 A: The proof does not look sound. Saying $\|f(x)-f(x_k)\|<\varepsilon'$ whenever $\|x-x_k\|<\delta'$ makes sense if $x_k\to x$. The tails of the sequence cannot be close to $x$ and $a$ at the same time unless $x=a$ (otherwise $\varepsilon''$ cannot be made arbitrary small).
The idea is more straightforward here: by definition


*

*Since $f$ continuous we have
$$
\forall \varepsilon>0\,\exists \color{red}\delta >0\text{ such that } \|f(\color{blue}x)-f(a)\|<\varepsilon,\text{ for all }\color{blue}x\colon \|\color{blue}x-a\|<\color{red}\delta.
$$

*Since $x_k\to a$ we have
$$
\forall\color{red}\delta>0\,\exists k_0 \in\mathbb{N}\text{ such that } \|\color{blue}{x_k}-a\|<\color{red}\delta,\text{ for all } k\ge k_0.
$$


We see that for any $\varepsilon>0$ we can find $\color{red}\delta>0$ in 1, then for this $\color{red}\delta$ find $k_0\in\Bbb N$ in 2. The tails of $\color{blue}{x_k}$ satisfy then $\|\color{blue}{x_k}-a\|<\color{red}\delta$, so we can take $\color{blue}x=\color{blue}x_k$ in 1 to conclude that for the tails $\|f(\color{blue}{x_k})-f(a)\|<\varepsilon$. We started from any $\varepsilon$ and ended up with the proper $k_0$, thus, obtained the sequential continuity by definition. 
A: Given $\epsilon>0$.
$f $ continuous at $a  \implies$
$\exists \delta_{a,\epsilon}>0$ such that $\|f (x)-f (a)\| $ whenever $\|x-a\|<\delta_{a,\epsilon}$
but $\lim x_n =a $ thus
$\exists N_0$ such that
$\|x_n-a\|<\delta_{a,\epsilon} $ whenever $n>N_0$.
finally
$n>N_0\implies \|x_n-a\|<\delta_{a,\epsilon}\implies \|f (x_n)-f (a)\|<\epsilon $
$\implies \lim_{+\infty}f(x_n)=f (a) $.
