I believe that there is a nice way to study this problem via the language of simplicial complexes and triangulations.
Let us triangulate $S^1$ and $S^2$ by the simplicial complexes $K$ and $L$, respectively, where $K$ consists of the proper faces of a $2$-simplex (i.e., a triangle) and $L$ consists of the proper faces of a $3$-simplex (i.e., a tetrahedron). If $h:S^1\to S^2$ is a continuous map, then we can apply the finite simplicial approximation theorem to conclude that there is a simplicial map $f:K\to L$ such that $h$ is homotopic to $f$. However, $f$ maps $K$ into the $1$-skeleton of $L$, and this is certainly a proper subspace of $L$. In particular, $h:S^1\to S^2$ is homotopic to a continuous function $f:S^1\to S^2 - p$ where $p$ is a point of $S^2$. Since $S^2 - p$ is homeomorphic to the contractible space $R^2$ (via stereographic projection), it follows that $h$ is homotopic to a constant map. Therefore, $\pi_1(S^2)=0$.
In fact, the technique I have used in the previous paragraph is more widely applicable and can be used to prove the following fact, which I leave as an exercise:
Exercise 1 Prove that the $i$th homotopy group of the $n$-sphere, $\pi_i(S^n)$, is trivial for all $0\leq i < n$ and all positive integers $n$.
I hope this helps!