A continuous map $f: S^1 \to S^2$ is homotopic to a point $f: S^1 \to S^2$ is a continuous map. Prove that $f$ is homotopic to a constant function.  Can someone give me a dircetion for proving this result? I think I should use the fact that $S^2$ is simply connected. I was able to prove the result given that $Im(f) \neq S^2$, but this is not generally true.

You can use the Lebesgue number lemma (though this may be overkill in the specific case of a sphere).
Specifically, take the open cover $\mathcal{O}=\{H^+,H^-\}$ of $S^2$ consisting of two thickened hemispheres. The Lebesgue lemma gives some number $\varepsilon$ such that any $\varepsilon$-ball in $S^2$ is contained in one or the other element of $\mathcal{O}$. Because $f$ is continuous on the compact set $S^1$, $f$ is uniformly continuous; choose $\delta$ such that $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$.
Now, divide $S^1$ into intervals $I_1,I_2,\dots,I_n$ of radius less than $\delta$, with midpoints $p_1,p_2,\dots,p_n$. Then $f(I_k)$ is always contained entirely within the $\varepsilon$-ball centered at $f(p_k)$, and hence within some element of $\mathcal{O}$. But any path in the thickened hemispheres $H^+$ and $H^-$ is homotopic (without moving the endpoints) to a great circle path, and so $f|_{I_k}$ is homotopic to a great circle path (without moving the endpoints).
It follows that $f$ is homotopic to some loop $g$ which is the union of finitely many great circle paths. But $g$ is nullhomotopic, as you've noted, and so $f$ is as well.
