"Nice" application of the fundamental group I'm looking for an example of a topological result which is easy to prove using the fundamental group, but hard or impossible elementarily.
First I thought about something like $\mathbb{R}^2\not\cong\mathbb{R}^n$ for $n>2$, but this should be possible to show without using the fundamental group but just simple connectedness.
Do you have any ideas? Thanks!
Edit: What about Brouwer's fixed point theorem; does it have an easy proof without using the fundamental group? If not, I think it would fit.
 A: Brouwer's fixed point theorem can be proved quite elementarily using Sperner's Lemma. 
The fundamental group can be used quite effectively to show certain topological spaces are not homotopic (and thus also not homeomorphic). For instance, none of the following spaces are homotopic: The torus, $\mathbb R^2$, projective plane.
With a little bit of the theory of covering spaces, one gets a very short and elegant proof that a free group on two generators contains subgroups that are free groups on any finite number of generators. This result is not terribly hard, but certainly not elementary in group theory. 
There is a nice proof of the fundamental theorem of algebra using the fundamental group (essentially using degree theory). 
A: I like the Phragmen-Brouwer Property (PBP). A topological space $X$ has the PBP if it is connected and the following holds: if $D,E$ are disjoint closed subsets of $X$ and $a,b$ are distinct points of $Y= X \backslash (D \cup E)$ which lie in the same component for both $X \backslash D$ and $X \backslash E$, then $a,b$ lie in the same component for $Y$.
Theorem If $X$ is path-connected and locally path connected, and the fundamental group of $X$ at any point does not have the integers $\mathbb Z$ as a group theoretic retract, then $X$ has the PBP. 
This is  9.2.1 of Topology and Groupoids.  Note that the circle $S^1$ does not have the PBP. I leave you to look at $S^n \backslash A$ for $n>1$ and $A$ a finite subset of $S^n$. 
(In older printings of T&G, the proof of 9.2.1 near the end refers to 9.1.9, instead of 8.4.1.) 
I suppose you could say this does not answer the question since the Theorem  involves the fundamental group, even if the PBP does not. However its application to particular examples does answer the question. 
Later: By the same methods, if  $X$ is path connected, and is the union of two open path connected sets whose intersection has $n$ path components, then the fundamental group of $X$ contains the free group on $n$ generators as a retract. So you can give examples of spaces which cannot be represented as a union in this way. 
Mind you, the $1$-dimensional Hurewicz Theorem would show you can get similar examples from using $H_1(X)$. 
A historical point is that the workers in topology of the early 20th century, such as Dehn, were aware of the importance of the nonabelian nature of the fundamental group in certain questions in analysis and geometry, and so were looking for higher dimensional nonabelian versions of the fundamental group. In 1932, E. Cech submitted a paper to the ICM at Zurich on higher homotopy groups, using maps of spheres $S^n$ for $n>1$; but Alexandroff and Hopf quickly proved these groups were abelian, and on these grounds persuaded Cech to withdraw his paper, so that only a brief summary appeared in the Proceedings, and Cech never worked again in that area. Of course Hurewicz, who is said to have been at that conference, did fundamental work on higherr  homotopy groups, and we now know that we can get (strict) nonabelian higher homotopy groupoids. 
A: 
Property: Let $f : \mathbb{D} \subset \mathbb{R}^2 \to \mathbb{D}$ be a continuous function such that $f(x)=x$ for all $x \in \partial \mathbb{D}$. Then $f$ is surjective.

This result can be prove by contradiction, using $f$ to get a retraction $\mathbb{D} \to \partial \mathbb{D}$ (the same argument can be found in the proof of Brouwer fixed point theorem). An alternative proof uses degree theory.
