Let $X$ be a topological vector space and $V$ be a linear subspace of $X$ such that $\text{dim}(X/V)=1$, then either V is closed or $\overline{V}=X$. In other words if it is not closed then it is dense.
I believe that this is true and TrialAndError's answer to "$\ker f$ is either dense or closed when $f$ is a linear functional on a normed linear space", seems to confirm this. But I would like to make sure that I have understood correctly why this is the case.
My reasoning is simply that if V is a subspace of X, so is $\overline{V}$. If V is not closed then they are distinct subspaces,and my intuition is that there isn't enough room for anything "bigger" than $\overline{V}$.
This is because: $\exists x_0 \in \overline{V} \setminus V$ and so $\text{span}\{x_0\} \subset \overline{V}$ and therefore so does $V+\text{span}\{x_0\}$ but $X=V\oplus\text{span}\{x_0\}$ (I guess I'm using the theorem of incomplete basis to say this) so $X=\overline{V}$. Perhaps there is a better way of seeing this without implicitly invoking the axiom of choice?
I'm not sure whether it's "better", but we can do it without implicitly using the axiom of choice.
Let $\pi \colon X \to X/V$ be the canonical projection. We know that $\overline{V}$ is a subspace of $X$ with $V \subset \overline{V}\subset X$. Then $W = \pi(\overline{V})$ is a subspace of $X/V$. But since $\dim X/V = 1$, it follows that either $W = \{0\}$, which means $\overline{V} = V$, or $W = X/V$, which means
$$X = \pi^{-1}(W) = \pi^{-1}(\pi(\overline{V})) = \overline{V} + V = \overline{V},$$
since generally for quotient spaces $X/Y$ and subspaces $Z$ of $X$ we have $\pi_Y^{-1}(\pi(Z)) = Z + Y$.