Nonconstant linear functional on a topological vector space is an open mapping

Consider arbitrary linear functional $\Lambda : X\mapsto \mathbb{R} $, pick $x^*\in X, \Lambda(x^*)=1$ (this can be done as we know $\Lambda$ is linear and $\exists x_0, \Lambda(x_0)=c\neq 0$, we can then pick $x^*=c^{-1}x_0$). We claim: $\forall V$ open subset of $X$, $\Lambda(V)$ is open in $\mathbb{R}$. Suppose $y\in \Lambda(V)$, we know $\exists x\in V, \Lambda(x)=y$. From the continuity of multiplication we know $\exists \delta>0, \forall |r| < \delta, x+ rx^* \in V, y+r=\Lambda(x+rx^*)\in f(V)$ and thus $B_\delta(y) = (y-\delta, y+\delta)\subseteq \Lambda(V) $. Therefore, $\Lambda(V)$ is open in $\mathbb{R}$ and $\Lambda$ is an open mapping.

Let $f : X \to \mathbb{R}$ be a nonconstant linear functional and $H= \text{ker}(f) \subsetneq X$. There exists an linear map $\varphi : X/H \to \mathbb{R}$ such that $f = \varphi \circ \pi$ with $\pi : X \to X/H$ the canonical projection. Using the definition of quotient topology, on can show that $f$ is open (resp. continuous) iff $\varphi$ is open (resp. continuous).

Two cases happen: either $f$ is continuous and $X/H$ is Hausdorff, or $f$ is not continuous and $X/H$ is not Hausdorff. However, any real one dimensional topological vector space is linearly homeomorphic to $\mathbb{R}$ if it is Hausdorff (see for example here), or it has the trivial topology otherwise.

To see this, let $Y$ be a real one dimensional topological vector space, say $Y= \mathbb{R}$, and $U$ be an open neighborhood of $0$. Suppose there exists $x \notin U$. Then, for all $y,z \in Y$ there exists $a \in \mathbb{R} \backslash \{0\}$ such that $y \notin z+aU$. Consequently, $Y$ is $T_1$ and the diagonal of $Y \times Y$ corresponds to $f^{-1}(0)$, with $f : (x,y) \mapsto x-y$, and it is closed for the product topology since $f$ is continuous; therefore, $Y$ is Hausdorff.

Consequently, either $X/H$ is Hausdorff, and $\varphi$ is a linear homeomorphism so $f$ is in fact continuous and open, or $X/H$ is not Hausdorff, and $\varphi$ is open (but not continuous) so $f$ is open (and not continuous).

Let $f$ be any non-constant functional on $X$ and $A$ to be any open subset in $X$.

$f$ is an open mapping if and only if for any $x\in A$, there exists an open subset $U$ of $0$ such that $x+U \in A$ and $f(x+U)=(f(x)-\epsilon, f(x)+\epsilon)$, for some $\epsilon>0$.

Since $f$ is a non-constant functional, according to Theorem 1.18 in Rudin's book, $f$ is bounded in some neighborhood of $0$. Take $W$ to be a neighborhood of $0$ such that $x+W \in A$. By Theorem 1.14, we can take a balanced neighborhood $U'$ of $0$ such that $U' \in W\cap V$. Since $f(U')$ is also a balanced neighborhood of $0$ in $\mathbb{R}$, $f(U')$ is a finite interval whose interior contains $0$. Take $\epsilon>0$ such that $(-\epsilon, +\epsilon)$ is contained in $f(U')$. $f^{-1}(-\epsilon,+\epsilon)= U$ is a neighborhood of $0$ in $X$ and $f(U)=(-\epsilon,+\epsilon)$. $U$ is the desired neighborhood.