Are these sets homeomorphic? 
Are $X_1 = \mathbb{R}^2 \setminus \{0\}$ and $X_2 = \{x \in \mathbb{R}^2: 0 < ||x||<1 \}$ homeomorphic? Is $X_2$ homeomorphic to $X_3 = \{x \in \mathbb{R}^2: 1 < ||x|| < 2 \}?$

I want to say that they are, but I'm having trouble writing down an explicit homeomorphism $f_1:X_1 \to X_2$ and $f_2:X_2 \to X_3.$ How do we construct functions $f_1,f_2$ between these sets such that $f_1,f_2$ are continuous, bijective and have continuous inverses? Any hints appreciated
 A: Hint: In terms or polar coordinate, you need only to find a homeomorphism
$$(0, \infty) \cong (0,1) \cong (1,2).$$ 
A: CW: This is merely expanding on John Ma's hint in case someone/someday/somewhere reads it and is not quite sure what is meant.
Focusing just on the first quadrant, viewed as a subspace of $\mathbb{R}$:
The open interval $(0,1)$ can be mapped to $(1,2)$ by the homeomorphism $x \mapsto x+1$.
(I'm guessing that wasn't where you struggled.)
Next, we seek a homeomorphism between the open interval $(0, \infty)$ and $(0, 1)$. 
How about mapping the former to the latter using the homeomorphism $x \mapsto \frac{x}{x+1}$?
Great; now spin around!
Concretely, to show, e.g., $X_1$ and $X_2$ are homeomorphic: 
Given a point in $X_1$, rotate it $\rho$ degrees onto the axis in the first quadrant; then map it to $(0,1)$ using $x \mapsto \frac{x}{x+1}$; then rotate it back by $-\rho$ degrees so that it lands comfortably in $X_2$. 
This composition of homeomorphisms gives a homeomorphism $X_1 \rightarrow X_2$, and hopefully the remaining details/verification are all well within reach.
