I expanded my comments into an answer.
In infinite dimensional normed linear spaces, subspaces are convex but not necessarily closed.
Consider $l_\infty(\mathbb{R})$ which is the set of bounded sequences in $\mathbb{R}$ with the norm $|(a_n)_{n \in \omega}| = \sup a_n$. Note that the vector space structure is given by term by term addition and term scalar multiplication.
Then let $M$ be the subspace of sequences that have only finitely many nonzero terms. You can verify that $M$ is a subspace.
The sequence $\alpha = (1, \frac{1}{2}, \frac{1}{3}, ..., \frac{1}{n}, ...)$ is a bounded sequence, hence is in $l_\infty(\mathbb{R})$. $\alpha \in \overline{M}$, the closure, since
$|\alpha - (1, \frac{1}{2}, ..., \frac{1}{n}, 0, 0, 0, ...)|_\infty = \frac{1}{n + 1}$.
So $\alpha$ is the limit of a sequence of elements from $M$. $\alpha$ is in the closure of $M$ but not an element of $M$. Hence $M$ is not closed.