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?

Suppose $0$ is not in the range of $h$. Then we have $h : S^2 \to \mathbb{R}^2 - \{0\}$.
Consider that the map $p_n : (\mathbb{R}^{n + 1} - \{0\}) \to S^{n}$ given by $p(x) = \frac{x}{|x|}$ is not only a retract, but also a homotopy equivalence for all $n \in \mathbb{N}$. So we have a continuous map $g = p_1 \circ h : S^2 \to S^1$ such that for all $x \in S^2$, $g(-x) = -g(x)$.
But this is not possible. For consider that $p : \mathbb{R} \to S^1$, with $p(x) = (\cos(x), \sin(x))$, is the universal cover of $S^1$. Consider the point $z = (1, 0, 0)$, and consider some $t$ such that $p(t) = g(z)$. Now since $S^2$ has fundamental group $0$ and $S^2$ is both path connected and locally path connected, and since $p$ is a universal cover, there exists a unique continuous $f : S^2 \to \mathbb{R}$ such that $f(z) = t$ and such that $g = p \circ f$.
Now consider $f(-z)$. We know that $f(-z) = -f(z)$. Therefore, we have $p(f(-z)) = -p(f(z))$. Therefore, $p(f(z)) = p(f(-z) + \pi)$. Now recall that for all $a, b \in \mathbb{R}$, $p(a) = p(b)$ if and only if $a = b + 2 \pi n$ for some $n \in \mathbb{Z}$. So we have $f(z) = f(-z) + \pi + 2 \pi n = f(-z) + \pi (2n + 1)$ for some $n \in \mathbb{N}$.
Now define $f_2 : S^2 \to \mathbb{R}$ by $f_2(x) = f(-x) + \pi (2n + 1)$.
We see that $f_2$ is continuous, and that $f_2(z) = f_2(-z) + \pi (2n + 1) = f(z) = t$.
And we see that for any $x$, $p(f_2(x)) = p(f(-x) + \pi (2n + 1)) = -p(f(-x)) = -g(-x) = g(x)$. That is, $p \circ f_2 = g$.
Therefore, $f_2 = f$, since $f$ is the unique continuous function such that $f(z) = t$ and $p \circ f_2 = g$.
Then we see that $f(-z) = f_2(-z) = f(z) + \pi (2n + 1) = f(-z) + \pi (2n + 1) + \pi(2n + 1)$. That is, $f(-z) = f(-z) + 2 \pi (2n + 1)$. That is, $2 \pi (2n + 1) = 0$. That is, $2n + 1 = 0$. That is, $n = -1/2$. But $n \in \mathbb{Z}$; contradiction.
Therefore, we see that $0$ must have been in the range of $h$.
This proof seems a little sterile, so let me briefly explain the "intuition" behind why $g : S^2 \to S^1$ with $g(x) = g(-x)$ cannot exist.
The idea is that we start at point $z$ and walk halfway around a circle to $-z$. Applying $g$ to this path gives us a path from $g(z)$ to $-g(z)$. This path must have a winding number of $k \pi$ where $k$ is odd.
But if we walk around the other half of the circle from $-z$ to $z$, we must be tracing out another path with exactly the same winding number $k \pi$. So the total winding number of the path is $2k \pi$. Since $k$ is odd, $2k \pi \neq 0$. But this means that the image of the complete circle must be a nontrivial loop, which is impossible since we can contract the circle, as a subset of $S^2$, to a single point on $S^2$.
