In Conway's book on functional analysis there is the following exercise:
If $\mathscr{X}$ is a topological vector space (TVS) and $\mathscr{M}$ is a closed linear space, then $\mathscr{X}/\mathscr{M}$ with the quotient topology is a TVS...
Now I wonder why we need that $\mathscr{M}$ is closed? Indeed can't we argue as follows:
We first show that $q:X\rightarrow X/M$ where $q(x) = x+M$ is an open map. For this reason we claim that $q^{-1}(q(U)) = U+M$ where $U$ is any subset of $X$. Indeed if $x\in q^{-1}(q(U))$ then $x+M= q(x)\in q(U)$. We can therefore find $y\in q(U)$ such that $x+M = q(y) = y+M$ but this implies that $x\in U+M$. Conversely if $x = y+m$ where $y\in U$ and $m\in M$ then \begin{equation*} q(x) = q(y+m) = y+M = q(y) \end{equation*} why $x\in q^{-1}(q(U))$. Since $X$ is a TVS $U+X$ is open whenever $U$ is open. Therefore $q$ is an open mapping in this case.
Consider $x_0+y_0+M$ and let $V$ be any open neighborhood containing this point. By the definition of the quotient topology $q^{-1}(V)$ is open and it contains $x_0+y_0$. Since $X$ is a topological vector space we can find neighborhoods $U_x$ and $U_y$ of $x_0$ and $y_0$ respectively such that $U_x+U_y\subseteq q^{-1}(V)$. By linearity of the quotient map we find that \begin{equation*} V\supseteq q(U_x+U_y) = q(U_x)+q(U_y). \end{equation*} Since $q$ is a open map $q(U_x)$ and $q(U_y)$ are both open in $X/M$ and therefore $q(U_x)\times q(U_y)$ is open in the product topology. This proves that addition is continuous in $X/M.$
Most books consider only Hausdorff topological vector spaces. [ I don't have Conway's book with me now]. Though your argument is correct you need closedness of $\mathcal M$ to make the quotient space Hausdorff.