How to show that the unit disk in $\mathbb{R}^2$ is equinumerous to $\mathbb{R}^2$? Prove that set $D=\{ ⟨x,y⟩ \in \mathbb{R}^2\,|\, x^2 + y^2 \leqslant 1 \}$ is equinumerous to $\mathbb{R}^2$. 
I know that $D \leqslant\mathbb{R}^2 $, since $D \subseteq \mathbb{R}^2 $, but I have no idea how to show that $D \geqslant\mathbb{R}^2 $. I was thinking about constructing a circle with radius $= 1$, and and center at $(0, 0)$, but then I got stuck. Could you please help me and explain how to solve this example? 
 A: Consider the map$$\begin{array}{ccc}\left\{(x,y)\in\mathbb{R}^2\,\middle|\,x^2+y^2<1\right\}&\longrightarrow&\mathbb{R}^2\\v&\mapsto&\dfrac v{1-\|v\|}.\end{array}$$It is a bijection. Since $\left\{(x,y)\in\mathbb{R}^2\,\middle|\,x^2+y^2<1\right\}\subset D$, this proves that $\#D\geqslant\#\mathbb{R}^2$.
A: 
Suppose that the closed unit disk $\overline{\mathbb{D}}^2$ in $\mathbb{R}^2$ is countable. Then we can construct a countable sequence of countable closed disks $\{n\times\overline{\mathbb{D}}^2\}_{n<\omega}$ such that $$\bigcup_{n<\omega}\{n\times\overline{\mathbb{D}}^2\}=\mathbb{R}^2,$$ thus $\mathbb{R}^2$ is countable since it is the countable union of countable sets. This is a contradiction since $\mathbb{R}$ is uncountable, thus we conclude that $\overline{\mathbb{D}}^2$ must be  uncountable.

EDIT: As dxiv mentions below, this proof only establishes that $\mathbb{R}^2$ and $\mathbb{D}^2$ are equinumerous if we are assuming that the Continuum hypothesis holds -- if we assume that there are cardinalities strictly between $\omega$ and $\mathfrak{c}$, this argument would require more work.
If we're assuming the continuum hypothesis it's often easier to work through contradiction when trying to establish the comparative cardinality of two sets, especially if one contains the other and we already know the cardinality of one of the two. When assuming that the continuum hypothesis is false, I believe that checking bijections as suggested in another answer is the fastest method if you can find one.
A: The map $x\mapsto\frac13\arctan x$ bijects $\mathbb R$ to the open interval $(-\frac\pi6,\frac\pi6),$ and so the map $$(x,y)\mapsto\left(\frac13\arctan x,\frac13\arctan y\right)$$ bijects the plane $\mathbb R\times\mathbb R$ to the open square $(-\frac\pi6,\frac\pi6)\times(-\frac\pi6,\frac\pi6)$ which is a subset of the unit disk.
A: Here's a visualizable way to biject the interior of the disk $D$ onto $\mathbb R^2$. Think of $\mathbb R^2$ as the $x$-$y$ plane in $\mathbb R^3$. Let $H$ be the open lower hemisphere of the sphere of radius 1 centered at $(0,0,1)$. So by projecting points straight down (parallel to the $z$-axis), you get a bijection between $H$ and the interior of $D$.  On the other hand. projecting radially from the center $(0,0,1)$ of the sphere you get a bijection between $H$ and the $x$-$y$ plans $\mathbb R^2$.  Compose the two bijections to get a bijection between the interior of $D$ and $\mathbb R^2$.
A: Do you know that a line segment, $\Bbb R$, and $\Bbb R^2$ are equinumerous?  If so, note that the segment $(0,0)$ to $(\frac 12,0)$ is a subset of $D$ and $\Bbb R^2$ is a superset of $D$.
