Let $A\subset \mathbb{R}^n$ be a set, $f:A\to\mathbb{R}^m$ be a function, $a\in A$. Suppose that $f$ is sequentially continuous at $a$. Prove that $f$ is continuous at $a$ in the sense of $\varepsilon-\delta$ definition.
My proof:
Let $\{x_k\}\subset A$ be a sequence such that $x_k$ converges to $a$. Then, since $f$ is sequentially continuous on $A$, $\forall \varepsilon >0$, $\exists k_0\in\mathbb{N}$ such that $\|f(x_k)-f(a)\|<\varepsilon$ whenever $k\ge k_0$. Also, $\forall\delta >0$, $\exists k_1\in\mathbb{N}$ such that $\|x_k-a\|<\delta$ whenever $k\ge k_1$. Let $k':=\max\{k_0, k_1\}$, let $k\ge k'$ (we can fix $k$), then $\|x_k-a\|<\delta\implies \|f(x_k)-f(a)\|<\varepsilon$. Hence, $f$ is continuous at $a$ in the $\varepsilon-\delta$ sense.
The marker deducted about 67% of the marks and commented that my argument is done for one sequence convergent to $a$ instead of covering all sequences convergent to $a$. However, I don't understand this argument because $x_k$ was taken to be an arbitrary sequence converging to $a$. So what is my mistake then?
Your mistake was to pick a sequence, then to pick $\varepsilon > 0$, then to pick $k_{0}$, then to pick $\delta > 0$ "satisfying the definition of continuity for your sequence".
Instead you need to show "there exists a single $\delta > 0$ that 'works' for every sequence". The natural approach here is contrapositive: Assume $f$ is not $\varepsilon$-$\delta$ continuous at some point $a$, and to construct a sequence $(x_{k}) \to a$ such that $(f(x_{k})) \not\to f(a)$, so that $f$ is not sequentially continuous at $a$.