Here a topological vector space is defined as a real or complex vector space equipped with a topology which is Hausdorff, and makes vector addition and scalar multiplication continuous.
Let $X$ be a real/complex topological vector space, and $M$ a subspace (that is not necessarily closed). Prove the quotient map $Q: X\to X/M$ is an open map.
My attempted proof is as follows. But I don't think I ever used the requirement that the topology is Hausdorff. Is this a problem? Or perhaps there can be a better proof? Thanks in advance.
Proof. Let $U\subset X$ be an open set. Suppose $Q^{-1}\left(Q(U)\right)=U+M$ is not open in $X$, then it contains at least one boundary point $u_0+m_0$, where $u_0\in U$ and $m_0\in M$. This means there exist two nets $(x_i)_{i\in I}\subset U+M$ and $(y_j)_{j\in J}\subset X\setminus (U+M)$ converging to this same boundary point: $$x_i \to u_0+m_0; \quad y_j\to u_0+m_0.$$ Then, since $X$ is a TVS, we also have net convergence $$x_i -m_0 \to u_0; \quad y_j-m_0\to u_0.$$ Note the second net completely lies outside of $U$, whereas the first net, by convergence, eventually lies in $U$. Point $u_0$ is then a boundary point of open set $U$. Contradiction. ${\rm \square}$
PS: At this point, we don't know whether $X/M$ is a TVS yet.
Update:
I saw another proof using the following lemma.
Lemma. Let $X$ be a TVS. Then, $U\subset X$ is a neighborhood of $0$ iff, for every $x\in X$, $x+U$ is a neighborhood of $x$.
Proof of lemma. Since $X$ is TVS, the translation for a fixed $x\in X$, $T_x: y\mapsto x+y$, is a homeomorphism from $X$ to itself. Hence the conclusion holds. $\square$
Proof of proposition. Let $U\subset X$ be an open subset. Then, by the lemma, $$Q^{-1}\left(Q(U)\right)=U+M=\bigcup_{x\in M}(x+U)$$ is a union of open sets, hence $Q(U)$ is open by the definition of quotient topology. $\square$.
But this proof still doesn't use Hausdorffness.