Are these subsets homeomorphic? Are the two subsets of the Euclidean Plane $[0,1]\times[0,1)$ and
$[0,1)\times[0,1)$ homeomorphic or not?
My attempt:
We need to find a bijective function $f$ from $[0,1]$ to $[0,1)$ such that $f$ and $f^{-1}$ are both continuous. If $f$ is continuous then for $U\subseteq [0,1)$, we have that $f^{-1}(U)$ is open in $[0,1]$ if $U$ is open in $[0,1)$. Using the identity map and any open subset of $[0,1)$ we see that the pre-image is open in $[0,1]$ so they are homeomorphic.
Does this make sense? 
 A: I will try to address this in somewhat detailed form and using elementary methods.
First, let us map the square $S$ (possibly with some boundary points removed) homeomorphically onto the disk $D$ (with corresponding boundary points removed). For convenience, I will take $S = [-1,1] \times [-1,1]$  and $D$ = unit disk, which is inscribed in $S$. We don't lose any generality by doing this since it is very easy to show that any two squares in the plane are homeomorphic (and similarly for disks). (Compose a dilation with a translation; these are both homeomorphisms, so their composition is also a homeomorphism.)
I will construct a map $f:D \to S$ piecewise, by cutting $D$ into the four cardinal quadrants delimited by the two lines $y = \pm x$. In the eastern quadrant, using polar coordinates $(r,\theta)$ (so that $-\pi/4 \leq \theta \leq \pi/4$), define $$f(r,\theta) = \left( \frac{r}{\cos \theta},\theta\right).$$ Note that $f$ is continuous.
For a fixed angle $\theta$, as we move from the center of the disk ($r=0$) to the boundary ($r=1$), the corresponding point on the square $S$ moves from the center of the square to the boundary of the square. On the western quadrant, we use the same formula except that the fist coordinate has a minus sign.
Analogously, on the northern quadrant, define $$f(r,\theta) = \left( \frac{r}{\sin \theta},\theta\right),$$ and finally add a minus sign to the first coordinate to define $f$ on the southern quadrant.
By the pasting lemma, these glue together to form a continuous map $f:D \to S$. Moreover, it is easily seen to be bijective (this was the point of the construction). As $D$ is compact and $S$ is Hausdorff, $f$ is a homeomorphism.
Note that $f$ maps the boundary of $D$ homeomorphically onto the boundary of $S$, so that it interacts well with the removal of boundary points from either space. Notably, the square $[-1,1] \times [-1,1)$ is mapped onto the disk with the boundary arc $\{\pi/4 \leq \theta \leq 3\pi/4\}$ removed, and the square $[-1,1) \times [-1,1)$ is mapped onto the disk with the boundary arc $\{-\pi/4 \leq \theta \leq 3\pi/4\}$ removed. The problem is reduced to showing that a disk with a compact, proper boundary arc of length $a$ removed is homeomorphic to a disk with a compact, proper boundary arc of length $b$ removed (for some $0 < a < b < 2\pi$).
Up to some rotation (which is a homeomorphism), we may assume that the removed arc on the first (resp. second) disk starts at $(1,0)$ and ends at $(1,a)$ (resp. $(1,b)$) (still in polar coordinates). Identify an automorphism $g$ of the circle with a continuous map $g:[0,2\pi] \to [0,2\pi]$ such that $g(0) \equiv g(2\pi) \mod 2\pi$. Then we may describe the automorphism $h:D \to D$ we seek through the following slight perturbation of the identity map, which restricts to a piecewise-linear map on each circle $\{r = \mathit{const}\}$:
$$h(r,\theta) = \begin{cases} \left(r,\frac{b}{a}\theta\right) & \text{if $0 \leq \theta \leq a$} \\ \left(r,\frac{2\pi-b}{2\pi-a}(\theta-2\pi)\right) & \text{if $a \leq \theta \leq 2\pi$} \end{cases}.$$
This fixes $(r,0)$ and sends $(r,a)$ to $(r,b)$; moreover, it is easily seen to be a homeomorphism. This concludes the proof.
A: They're homeomorphic since both of them are homeomorphic to the closed unit disc in $R^2$  with an arc removed on its boundary.
