Why should K be closed to ensure X/K is complete? If $K$ is a closed subspace of Banach space $X$, then $X/K$ is complete. But I think the usual proof of this theorem doesn't make use of the fact that $K$ is closed. Would anyone explain it to me? Thanks a lot.
 A: If the subspace $K$ is not closed, then the quotient $X/K$ is not even Hausdorff, so does not meet the usual requirements of a topological vector space at all!
(It's not about completeness or not, as $X/K$ is not metric in that case!)
EDIT: In more detail, for $K$ not closed, and for $x$ in the closure of $K$ but not in $K$ itself, every neighborhood of $0$ in the quotient (with the quotient topology) contains $x$, but $x\not=0$ in the quotient. So the quotient would be non-Hausdorff (since two distinct points, $0$ and $x$, do not have disjoint neighborhoods).
A Banach space, or even a normed space, is Hausdorff, as is every metric space.
A: If $K$ is not closed, then the function $\lVert x+K\rVert=\inf_{y\in K} \lVert x-y\rVert$ is not a norm on the quotient space, but just a seminorm, because $\lVert x+K\rVert=0$ for all $x\in\overline K$.
A: You do need closure of $K$. In particular, you need it to show that $X/K$ is in fact a normed space: If $\| x + K \|=0$, closure of $K$ implies that $x+K=K=0_{X/K}$. 
