Let $A=\{\alpha=(a_1,a_2,...):a_n\in\mathbb{C},\lim_{n\rightarrow \infty}a_n\text{ exists }\}$ the set of convergent complex sequences and $B=\{\alpha=(a_n)\in A:\lim_{n\rightarrow \infty}a_n=0\}$ the set of complex sequences converging to zero.
How do I show that $B$ is a closed linear subspace of $A$?
I think I have to show that $A\backslash B$ is open by constructing an open ball around any sequence in $B$. But I wouldn't know how to do that.
It is certainly easier to show that a convergent sequence of elements from $B$ has its limit also in $B$; this is an equivalent definition of closed sets in a metric space. Remember that you are taking sequences of sequences here.
Edit: (Sketch of proof). Let $(a_n^{(k)})_{k=1}^\infty$ be a sequence in $B=c_0$, and suppose that $(a_n^{(k)})\to (a_n)\in A$ as $k\to\infty$. (That is, the sequence $(a_n^{(k)})_{k=1}^\infty$ converges in $A$ to $(a_n)$. In particular $a_n^{(k)}\to a_n$ for each $n\in\mathbb N$, and this convergence is uniform over all $n$.) Set $a=\lim_{n\to\infty} a_n$ which exists because $(a_n)\in A$. The proof is complete by showing that $a=0$. Let $k,n\in\mathbb N$ and look at $$ |a|=|a-a_n+a_n-a_n^{(k)}+a_n^{(k)}|\leq |a-a_n|+|a_n^{(k)}-a_n|+|a_n^{(k)}|. $$ It remains to show that the right hand side of the above can be made small for all sufficiently large $k,n$, which I leave to you.