According to my notes
$K\subset\mathbb{C}$ sequentially compact $\iff$ Every sequence in K has a subsequence which converges to a Point in K
$A\subseteq\mathbb{C}$ closed set $\iff ((a_n)_{n\in\mathbb{N}}$ convergent sequence in A $\Rightarrow \lim_{n\rightarrow \infty}a_n\in A$
I am wondering why the definitions, save for $\mathbb{C}$ are not equivalent
If $A\neq\mathbb{C}$ and $A$ closed set, why does $A$ not have to be necessarily compact? Because the opposite is true. If I take a convergent sequence of a sequentially compact subset $K$ then the convergencepoint has to be by Definition in $K$.
Take $A=\{z\in\mathbb{C}\,|\,\lvert z\rvert\geqslant1\}$, for instance. It is closed. However, it is not sequentially compact because, for instance, the sequence $1,2,3,\ldots$ has no convergente subsequence. All you can deduce about sequences of elements of $A$ from the fact that $A$ is closed is that if one such sequence converges, then its limit belongs to $A$.