Neither open nor closed subspace of a vector space?

For a different flavor of example, consider $C[0,1]$ with the uniform norm. By a standard theorem (Stone-Weierstrass), the space $P$ of polynomial functions is dense in this space. But then, since $P$ is dense but clearly not equal to the whole space, it can't be closed in it.

Most of the standard Banach spaces you know are separable - which, in practice, means that they have a "nice" dense subspace. That dense subspace is a non-closed subspace.

You have to look to infinite dimensional spaces. In finite dimensional spaces, a subspace is always closed.

An example for an infinite dimensional space

There is a theorem stating that a linear form $\varphi$ is continuous if and only if its kernel $W$ is a closed subspace. And if $\varphi$ is discontinuous then the kernel is dense in $V$.

Hence if we find a non-continuous linear form, its kernel cannot be closed according to theorem above. And cannot be open either as an open subspace that is dense is the space $V$ itself.

Take $V=\{P \in \mathcal C([0,1]), \mathbb C) \mid P \mbox{ is a polynomial}\}$, equipped with the $\sup$ norm. The linear form $\varphi: P \mapsto P(3)$ is discontinuous. To see this consider the sequence $P_n(x) = (x/2)^n$. The sequence converges to zero for the $\sup$ norm but $\lim\limits_{n \to \infty} \varphi(P_n) = \infty$.

The kernel $W$ of $\varphi$ is the strange animal you're looking for!