The fundamental group of the Hawaiian earring is very complicated, but since it's "1dimensional" one might guess that the higher homotopy groups vanish. Do they? Since the Hawaiian earring does not have a universal cover, the standard approach to showing that higher homotopy groups of graphs vanish does not apply.
3 Answers
Despite the nonexistence of a universal covering space, there is still an object that acts like a universal covering space for the earring space. It is actually quite useful for understanding the structure of the fundamental group as a subgroup of the inverse limit $\varprojlim F_n$ of free groups. The "generalized universal covering" still provides all the same unique path lifting properties for maps out of locally path connected spaces that one might want. The study of these particular coverings for wild spaces is mainly due to Hanspeter Fischer and Andreas Zastrow. I think you should be able to use this to mimic the argument for graphs.
Let $\mathbb{E}$ be the earring space, which is a shrinking wedge of a sequence of circles. Let $\widetilde{\mathbb{E}}$ be the set of homotopy classes of paths starting at the basepoint. Take $p:\widetilde{\mathbb{E}}\to \mathbb{E}$, $p([\alpha])=\alpha(1)$ to be the endpoint projection and give $\widetilde{\mathbb{E}}$ the usual "whisker" topology that you do to construct universal covers: the basic neighborhoods are $B([\alpha],U)=\{[\alpha\cdot\epsilon]\epsilon([0,1])\subset U\}$ where $U$ is open in $\mathbb{E}$. It turns out that $p:\widetilde{\mathbb{E}}\to \mathbb{E}$ is an open map which has unique lifting of all paths and homotopies of paths. Moreover, if $Y$ is path connected, locally path connected, and simply connected, then any based map $f:Y\to \mathbb{E}$ has a unique lift $\tilde{f}:Y\to \widetilde{\mathbb{E}}$ such that $p\tilde{f}=f$. Consequently, $p$ induces isomorphisms on higher homotopy groups.
It also turns out that $\widetilde{\mathbb{E}}$ has the structure of an $\mathbb{R}$tree (uniquely arcwise connected metric space where each arc is isometric to a subarc of the reals), and these are known to be contractible (this is due to J. Morgan I believe).
Now you should have all the usual ingredients to mimic the usual argument for graphs. In fact, the earring is not special here. This should work out for all 1dimensional Peano continua.

2$\begingroup$ I guess I never appreciated the importance of the condition "locally simply connected" in the construction of the universal cover. $\endgroup$ Commented Apr 21, 2014 at 0:48

1$\begingroup$ @LiorSilberman : $\:$ I thought that standard facts about covering spaces only $\hspace{1.67 in}$ needed "semilocally simply connected". $\;\;\;\;$ $\endgroup$– user5810Commented Apr 21, 2014 at 3:00

I did not read this paper, but it claims that the answer is yes: Onedimensional sets and planar sets are aspherical by J.W. Cannon, G.R. Conner and Andreas Zastrow.

5$\begingroup$ Also available publicly via an author's website $\endgroup$ Commented Apr 20, 2014 at 1:46

4$\begingroup$ It is actually available publicly from the publisher's website. $\endgroup$ Commented Apr 20, 2014 at 11:02

1$\begingroup$ @AlexeyMuranov for now... $\endgroup$– David Roberts ♦Commented Jun 25, 2017 at 3:22

$\begingroup$ If I understand correctly Paul's comments to his answer to this very post (mathoverflow.net/a/163868/14094) there is a gap in this general result that (general) compact sets are aspherical. $\endgroup$– YCorCommented Jun 25, 2017 at 8:54

2$\begingroup$ The paper of Cannon/Conner/Zastrow correctly proves planar compacta are aspherical. My answer/comments below sketches an alternate, strategy/technique to achieve the same result via different methods. I expect the gaps can be filled in, but the purported alternate proof has yet to be written up carefully and published. $\endgroup$ Commented Jun 25, 2017 at 13:46
Here is a short natural argument that planar continua are aspherical, different from the technique of Cannon/Conner/Zastrow, and straight forwardly applied to the Hawaiian earring.
The Hawaiian earring (and more generally any planar compactum $X$) is the nested intersection of planar polyhedra $X_n$.
The planar Euclidean metric naturally induces a length structure on $X_n$ so that $X_n$ is locally CAT(0).
In particular there is a natural proof $X_n$ is aspherical, since inessential loops in $X_n$ can be canonically extended to maps of the disk, and since the mball $B_{m}$ is naturally fibred by disks, so that the boundaries of the disks fibre the msphere $S_{m}$.
To see $X$ is aspherical take a map $f: S_{m} \rightarrow X$, and obtain the natural extensions $F_{n}: B_{m} \rightarrow X_{n}$.
Ascoli's theorem ensures the existence of subsequential limit and hence $X$ is aspherical.
In Cannon/Conner/Zastrow, inessential loops in $X$ are contracted internally, within their own image. In contrast, in the argument at hand, roughly speaking, inessential loops in $X$ bound disks fibred by limits of external geodesic chords.
To apply Ascoli's theorem we must gain global control of the equicontinuity data.
It is useful to show there exists a universal constant K>0 so that if P is a planar polyhedron, if $f: [0,1] \rightarrow P$ is a path, if C is the local geodesic path homotopic to f, (and if g is the parameterization of C induced by projection of f), then if (d,e) is uniform continuity data for f with respect to absolute value, then (d,Ke) is uniform continuity data for g (also with respect to absolute value). K=1.5 is apparently the worst case.
However, in the special case of the Hawaiian earring we may use K=1, since $X_n$ is the union of a finite bouquet of loops and a convex disk.

$\begingroup$ Nice argument; so you claim that in general planar continua (i.e., not for Hawaiian earring) there is a gap; can you fix it? $\endgroup$ Commented Apr 21, 2014 at 0:18

$\begingroup$ There is a genuine gap since K=1.5 can actually happen. Place the letter V in the upper half plane with vertex at (0,0) and angles 60 degrees with w.r.t. x axis. Now place a different V with origin vertex, with angles 30 degrees and very long edges. Now project orthogonally the first V into the 2nd V while fixing the origin. This is purportedly the worst case, yielding a kind of poor man's GehringHayman Theorem. $\endgroup$ Commented Apr 21, 2014 at 4:37

$\begingroup$ It is more straightforward to see K=3 is a universal upper bound. Let $Q$ denote the universal cover of $P$. Note $Q$ is CAT(0). Take any two geodesics segments A and B in $Q$. In $Q$, naturally project A into B. Call the image C. Now map A and C naturally into the plane and compare the ratios of the their diameters, using absolute value. $\endgroup$ Commented Apr 21, 2014 at 14:19

$\begingroup$ If A and C are disjoint, K =1 suffices, and this is provable by induction on the number of segments comprising A and B. If A and C are not disjoint, let D denote the intersection. Let A=A1*DA2 and C=C1*DC2. The diameter of the planar projection of Ai is at most that of Ci. Hence K=3 is an upper bound on the ratio of diameters of the planar projections of A and C. $\endgroup$ Commented Apr 21, 2014 at 14:29