Quotient map is closed Let $X = [0,1]\times[0,1]$ and let $\sim$ be the equivalence relation for the Klein bottle. Let $q: X \to X/\sim$.
Let $C$ be closed in $X$. Then there is closed set $V$ in $\mathbb R^2$ so that $C = V \cap X$. 
How to show $q(C)$ is closed in $X/\sim$? 
 A: If you'd like to show this completely hands-on, here's what you can do. For every closed set $C$ consider set $D := q^{-1}(q(C))$. It is enough to show $D$ is closed because $q(C) = q(D)$ and $q(D)$ is closed precisely when $D$ is (by the definition of quotient topology).
Now, how does $D$ look like? For every point that $C$ shares with the boundary of $X$ you get another point on the opposite (as per the quotient) side. Also, if one of the corners belong to $C$ then all of the corners belong to $D$.
With this picture in mind, let $V$ be a closed set such that $V \cap X = C$. Write $W = V \setminus X^0$. By reflecting, rotating and translating $W$ a few times (in accord with the Klein-bottle quotienting) and taking union over the resulting sets (and $V$), we obtain a new closed set that intersects $X$ precisely in $D$. Try drawing the picture.
A: Define the following closed subsets of the square $M$: The four corner points $K$, the top boundary $T \cong [0,1]$, the bottom boundary $B\cong [0,1]$, the left boundary $L\cong [0,1]$ and the right boundary $R\cong [0,1]$. Note that reflection at either the line $y=\frac{1}{2}$ or $x = \frac{1}{2}$ is a homeomorphism. If $C$ is closed in $M$ then $D$ is either equal to $C$ or it is equal to $C$ union a reflected copy of $C$ intersected with one of the closed sets $T,B,L,R,K$. Hence $D$ is a union of closed sets and hence closed.
