Universal cover of $\mathbb R^2\setminus\{0\}$

Hint: Think of the mapping $z\mapsto e^z$ from $\Bbb{C}$ to $\Bbb{C}\setminus\{0\}$. Its derivative is also $e^z$, which is always non-zero. Therefore the mapping is conformal everywhere, i.e. a local homeomorphism.

One (of many) possibilities to consider is as for all infinite cyclic covers of nice $X$ is to pull back the universal cover $\mathbb R \to S^1$ of some $f: X \to S^1$. In the smooth setting $f^{-1}(1)$ (let 1 be a regular value) is a 1-codimensional submanifold and the infinite cyclic cover is obtained by cutting along that submanifold and gluing $\mathbb Z$-many copies together.

In your case the map is the projection $X=\mathbb R^2 -0 \to S^1$. And the submanifold is a ray of a point $\mathbb R_+ x$, canonically $x=1$. So take $\sqcup_{\mathbb Z} ( \mathbb R^2 - \mathbb R_{\geq 0})$ and cross-glue them together along one boundary component $\mathbb R _{>0}$.

For more info cf. e.g. my answer here.

Edit: to explicitely write down what I wrote above you can write $$\mathbb R^2 -0 = \mathbb R_{>0} \times S^1 \stackrel{p_2} \to S^1,$$ where the pull back becomes obvious: $$\begin{array}{c}p^*(\mathbb R) = \mathbb R_{>0} \times \overbrace{id^*\mathbb R}^{\mathbb R} &\to &\mathbb R\\ \downarrow && \downarrow \\ \mathbb R_{>0}\times S^1 &\stackrel {p} \to & S^1\end{array} $$ and the covering map is giving by $id \times \pi$ where $\pi:\mathbb R \to S^1$. By using the isomorphism $\mathbb R_{>0}\cong \mathbb R$ you get the universal cover:

$$ \mathbb R^2 \cong \qquad\mathbb R_{>0} \times \mathbb R \stackrel {1\times\pi} \longrightarrow \mathbb R_{>0}\times S^1 \qquad \cong \mathbb R^2 -0.$$