How can I prove that the hawaiian earring has no universal cover? I know that the Hawaiian earring is not semi-locally simply connected so the existence is not guaranteed. Also, the point in which it must fail is the origin, where it isn't even locally simply connected.
My first thought was to use that the universal cover must be simply connected and locally path connected (since the Hawaiian earring is locally path connected), but this does not imply that the universal cover is locally simply connected, which was my intention, so I don't think this is a good approach.
Does anyone know how could I solve this?
 A: If $X$ is the Hawaiian earring, assume universal cover $\widetilde{X}$ exists. Let $p$ be the "interesting point" of $X$ (namely, the origin). $q$ be the lift of $p$ in $\widetilde{X}$. If $U$ is an evenly covered neighborhood of $p$, then it lifts to a neighborhood $V$ of $q$ such that the projection $U \to V$ by restricting the covering map is a homeomorphism. 
$i : U \hookrightarrow X$ be the inclusion map. We can look at the composition $\pi_1(V) \stackrel{\cong}{\to} \pi_1(U) \stackrel{\pi_1(i)}{\to} \pi_1(X)$ where first map is the induced map of projection. This map is the same as $\pi_1(V) \stackrel{\pi_1(j)}{\to} \pi_1(\widetilde{X}) \to \pi_1(X)$ where $j$ is the inclusion $V \hookrightarrow \widetilde{X}$. 
But as $\widetilde{X}$ is simply connected, the composition is the zero map, hence so is $\pi_1(i)$. Thus, $U$ is a neighborhood of $p$ which makes $X$ semilocally simply connected at $p$. Contradiction, as every neighborhood of $p$ contains a small circle, and a loop going around that small circle can never be nullhomotoped in $X$.
