Let $X$ be a normed space. If $M$ is a subspace, then $X/M$ has a known seminorm:
$\left\|{x+M}\right\|=\inf\{\left\|{x+y}\right\|:y\in M\}$
It is easy to show that if $M$ is closed then $\left\|{}\right\|$ is a norm in $X/M$ (the only class with "seminorm" zero is the trivial class).
My question is: if $\left\|{}\right\|$ is a norm in $X/M$, can we prove $M$ must be closed?
I couldn't prove it so far. Is there a counterexample?
If $M$ is not closed, then there exists a convergent sequence $\left\{x_n\right\}_{n\in\mathbb{N}}$ of elements of $M$ with limit $x$ not in $M$. Thus, $x+M\neq M$ but $\|x+M\|_{X/M}=0$, so that $\|\_\|_{X/M}$ is not a norm on $X/M$.
Here is an alternative proof. Suppose that the proposed function $\lVert\:\cdot\:\rVert_{X/M}$ defines a norm on $X/M$. Then the quotient map $X \to X/M$ given by $x \mapsto x + M$ is a continuous linear map between normed spaces, since we have $$ \lVert x + M \rVert_{X/M} = \inf\big\{ \lVert x + y\rVert_X \: : \: y\in M\big\} \leq \lVert x + 0\rVert_X = \lVert x\rVert_X. $$ Now $M$ is the inverse image of a closed set (namely $\{0\}$) under a continuous map, which must therefore be closed.
