Show that $h:\mathbb{S^2} \to \mathbb{R}$ has a zero. 
Let do the usual, define $h:S^2 \to R$, by $h(x)=f(x)-g(x)$, then $h$ is continuous and clearly, $h(-x)=-h(x)$, so we have to find a zero for $h$.
As $S^2$ is connected, $h$ is continuous and for $x \in S^2$, if $h(x)>0$ then $h(x)=h(-(-x))=-h(-x)$ $\to$ $h(-x)<0$
We conclude that $h(S^2)$ is connected in $\mathbb{R}$ and $0\in h(S^2)$, so  by the intermediate value theorem there existe some $x_0 \in S^2$ such that $h(x_0)=0$.
Is this okay?
I thought of using the the Brouwer Fixed Point Theorem, but the domain is not $D^2$ as in the theorem, or there is a way to used it? Or some other tool of Algebraic Topology to do it?

As noted in comments, you haven’t shown $f(x_0)=g(x_0)=0,$ only that $f(x_0)=g(x_0).$
Assume there is no such $x.$ Then define $k: S^2\to S^1:$ $$k(x)=\frac{(f(x),g(x))}{\|(f(x),g(x))\|}$$
Then $$k(-x)=-k(x).\tag1$$
Show there can be no such $k:S^2\to S^1$ which satisfies (1).
Hint: Show there is a function $\alpha:S^1\to S^2$ such that $k\circ \alpha$ is not null-homotopic.
