Only the subspaces mentioned by the OP are obviously closed.

Let $V$ be a real vector space and $W\subset V$ a proper infinite-dimensional linear subspace. We shall endow $V$ with a norm so that $W$ will not be closed in $(V, \|\;.\;\|)$. For this we consider a Hamel basis for $W$ that we partition into a countable part $\{b_1, b_2, \dots\}$ and some (possibly empty) rest $B_1$. We augment this basis to a basis $B$ for $V$ by adding a distinguished vector $b_0$ and a (possibly empty) set $B_2$ (since $V\neq W$ there is such a $b_0$). For an element $x=\sum_{b\in B} b'(x)b$ (the $b'$ are the corresponding coefficient functionals) we put
$$
\|x\| = \sup_{b\in B_1} |b'(x)| + \sup_{b\in B_2} |b'(x)| + \sup_n |b_n'(x) + 2^{-n} b_0'(x)|.
$$
(This imitates the sup norm on the linear span of $c_{00}$ and $(2^{-n})$.)

Let $x_N= \sum_{k=1}^N 2^{-k}b_k$; we shall argue that $x_N\to b_0$. Since $x_N\in W$ and $b_0\notin W$ this shows that $W$ is not closed. Now
$$
\|x_N-b_0\| = \sup_n |b_n'(x_N-b_0) +2^{-n} b_0'(x_N-b_0)|,
$$
and for $n\le N$ this term is $=0$, whereas it is $=2^{-n}$ for $n> N$; consequently $\|x_N-b_0\|\le 2^{-(N+1)} \to0$.