Hint: To show that the set of sequences with $l^2$ norm less than or equal to $1$ is convex, let $x_n,n\in \mathbb{N}$ and $y_n,n\in \mathbb{N}$ be sequences with $l^2$ norm less than or equal to $1$ and observe that $(tx_i+(1-t)y_i)^2 \leq tx_i^2+(1-t)y_i^2$ for $t\in [0,1]$, hence the $l^2$ norm of the sequence $x+y$ is at most $t\|x\|_2+(1-t)\|y\|_2$.
As Robert said in his comment, this set is a convex body as a subset of $l^2$, so you need to tell us what vector space you're working over. If forced to guess, I would say you're probably working over $l^\infty$, as any square-summable sequence in $l^\infty$ plus a sequence $\epsilon,\epsilon,\ldots\in l^\infty$, which has $l^\infty$ norm $\epsilon$, is not square-summable. But again, you could clear this up if you tell us what vector space you're working over.