I have a topological vector space $E$ (not necessarily Hausdorff separated) and $p:E\to\mathbb{R}$ a linear continuous functional such that $p\neq0$ (not identically equal to zero). Why is $p$ an open map?
I see that, since $p\neq0$, $p$ is surjective but $E$ is not a Banach space to use the open mapping theorem.
Therefore there should be an easier argument.
Since $p$ is continuous, $Ker\ p:=p^{-1}(\{0\})$ is closed in $E$. The quotient space $E/Ker\ p$ is a $1-$dimensional Hausdorff separated topological vector space, algebraically isomorphic to $\mathbb{R}$ via $\hat{p}:E/Ker\ p\to\mathbb{R}$, $\hat{p}(e+Ker\ p):=p(e)$, $e\in E$.
We have the quotient map $\pi:E\to E/Ker\ p$, $\pi(x):=x+Ker\ p$ which is linear continuous and open (even though $E$ is not separated)!
Also $p=\hat{p}\circ \pi$, $p(D)=\hat{p}(\pi(D))$, for $D\subset E$.
Hence it suffices to note that $\hat{p}$ is an open map which can be proven directly because there is only one separated linear topology on $\mathbb{R}$ (the usual one) or by using the open mapping theorem for Banach spaces.