Mapping homotopic to the identity map has a fixed point

Have you seen homology and degree of continuous mappings? If so, this is pretty short:

Since $f$ is homotopic to the identity, $\deg f = \deg \text{id}_{S^2} = +1$. On the other hand, if a continuous map $g:S^n\to S^n$ has no fixed points, then $\deg g = (-1)^{n+1}$. This shows that $f$ must have a fixed point, since otherwise $\deg f = (-1)^{2+1} = -1$.

Alternatively, you can prove that any map $S^n\to S^n$ with no fixed point is homotopic to the antipodal map. (This is used to prove the statement about $g$ in the previous proof.) Note that if $f$ has no fixed points, then $$h(x, t) = (1-t)f(x)-tx$$ is nonzero, so $h(x,t)/|h(x,t)|$ defines a homotopy from $f$ to the antipodal map. (This proof is in Hatcher, Algebraic Topology, section 2.2.) And the antipodal map is homotopic to the identity if and only if $n$ is odd, so this proves the claim by contraposition.

Alternatively, this follows from the Lefschetz fixed point theorem. As $\phi$ is homotopic to the identity map, the Lefschetz number of $\phi$ is just the Euler characteristic of $S^2$, which is 2. As $2 \neq 0$, it follows that $\phi$ has at least one fixed point.