Let $B=\{(x,y)\in \mathbb R^2\mid x^2+y^2\leq 1\}$ and let $K$ be a compact subset of $X$. Suppose $\pi:\mathbb R^2\setminus B\to X\setminus K$ is a homeomorphism. Show that $X$ is homeomorphic to $\mathbb R^2$.
Conclusions I can make now are as follows:
-The fundamental group is a topological invariant so this means $\pi_1(\mathbb R^2\setminus B)\cong \pi_1(X\setminus K)$
-$S_1$ is a deformation retract of $\mathbb R^2\setminus B$ $\Rightarrow$ $\pi_1(\mathbb R^2\setminus B)\cong \pi_1(S^1)\cong \mathbb Z$
-Since $\mathbb R^2$ is simply connected, each fiber of $\pi$ has the same cardinality as the fundamental group of $X$.
It should be enough to show that $X$ is simply connected, but I'm not sure how.
Hint: Suppose that the inverse image of every point has at least two elements. At least one of these must lie in the unit ball. Choose a countable sequence that has no limit points in $X\setminus K$. Looking at the preimages in $B$ and the "evenly covered" property gives a contradiction.