Covering map homeomorphism? I have the following question:

Let $X$ be a topological space and $q: \mathbb{R}^2 \to X$ be a covering map. Let $B$ denote the closed ball of radius 1 centered at the origin and let $K \subset \mathbb{R}^2$. Suppose $q:\mathbb{R}^2\backslash B \to X\backslash K$ is a homeomorphism. Show that $X$ is homeomorphic to $\mathbb{R}^2$. 

One thing I was thinking is to show that $X$ is simply connected, then by Show that if B is simply-connected, then p is a homeomorphism. we have the desired result. But, I am not even sure if this is the right approach.
Any help would be greatly appreciated.
Thank You
 A: Let us prove a more general result:
Let $q :E \to B$ be a covering map, where $E$ is a connected, locally compact and non-compact $T_1$-space. Let $C \subset E$ be compact and the restriction $q : E \setminus C \to B$ be injective. Then $q$ is a homeomorphism.
Note that local compactness and compactness do not involve the Hausdorff property. Locally compact means that for each point and each open neighborhood $U$ there exists an open neighborhood $V$ and a compact $K$ such that $V \subset K \subset U$. 
Here are some facts for a map $f : X \to Y$ which is a surjective local homeomorphism (for example a covering map). 


*

*$X$ is locally compact iff $Y$ is locally compact. 

*$X$ is $T_1$ iff $Y$ is $T_1$.
Let us now prove our above theorem. Since $E$ is connected, so is $B$ and we see all fibers have the same cardinality. In fact, this cardinality is finite. Let $F$ be a fiber (which is closed since $Y$ is $T_1$). The intersection with $C$ is compact and discrete, thus finite. $F$ contains at most one additional point which then lies in $E \setminus C$.
Now assume that $B$ is compact. Using local compactness and finiteness of fibers we easily that $E$ is a finite union of compact sets, thus itself compact which is a contradiction. Hence $B$ is non-compact and we find a point $x$ not belonging to $q(C) $. The fiber over this point contains a single point which belongs to $E \setminus C$. This shows that our covering map has trivial fibers and thus is a homeomorphism. 
A: Here is another solution that I am aware of:
Since any covering map is surjective and open, it suffices to show that $q$ is injective. First, notice that $q^{-1}(K) \subseteq B$ and $K$ is closed (because $q$ is an open map and so $q(\mathbb{R}^2\backslash B) = X\backslash K$ is open in $X$). Being a closed subset of a compact set, $q^{-1}(K)$ is compact. Hence, $q|_{q^{-1}(K)}:q^{-1}(K)\longrightarrow K$ is a finite-sheeted covering map (see For a compact covering space, the fibres of the covering map are finite. for this result). This means that $|q^{-1}(\{k\})| = n$ for each $k \in K$. Moreover, by the result in $k$-fold covering of a connected set, $B$. $|q^{-1}(\{x\})| = n$ for each $x \in X$.  
Now, we claim that $q|_{B}:B\longrightarrow X$ is not onto. For if it were, then $X$ would be compact and since the covering map is finite sheeted, by the result in Finitely sheeted covering space of a compact space is is compact we would have $\mathbb{R}^2$ is compact, a contradiction. Hence, $q|_{B}$ is not onto. Thus, there is some $x_0 \in X$ such that $q(b) \neq x_0$ for any $b \in B$. Hence, $q^{-1}(x_0) \subseteq \mathbb{R}^2\backslash B$. Since $q|_{\mathbb{R}^2\backslash B}$ is a homeomorphism, in particular it is injective, we have that $|q^{-1}(\{x_0\})| = 1$. Hence, $|q^{-1}(x)| = 1$ for all $x \in X$ and so $q$ is injective.  
