Let g: $RP^2 \mapsto RP^2$ be a map such that $g_*$ the induced map on the fundamental group is not the zero map.
We need to prove that g can be lifted to a map $T:S^2 \mapsto S^2$ such that $T(-x)= -T(x) $ for all $x \in S^2$.
This is my approach: let $p$ the covering space from $S^2 $ to $RP^2$. Then by the lifting criterion there exists $h: S^2 \mapsto S^2$ such that $ p \circ h = g \circ p $. For $x \in S^2$, $p \circ h(x)= p \circ h(-x)$ hence $h(x)=h(-x)$ or $h(x)=-h(-x)$. If I am able to show that the first option is impossible we will be done.
To do that we can consider the loop $\gamma$ in $RP^2$ at $y$ where $y=p(x)=p(-x)$ that is lifted by the path from $x $ to $-x$ in $S^2$. Since $g_{*}$ is not the trivial map then $g_{*}$ takes $\gamma$ to $\gamma$. If $h(x)=h(-x)$ this means that $\gamma$ is lifted by a loop to $S^2$? Is this a contradiction? If so, why? Also, if there is better way to write this I would be grateful if you can show it to me.
It is a little more subtle. You say that $g : \mathbb RP^2 \to \mathbb RP^2$ is a map such that the induced map on the fundamental group is not the zero map. This means that we have a basepoint $y \in \mathbb RP^2$ such that $g(y) = y$ and $g_* : \pi_1(\mathbb RP^2,y) \to \pi_1(\mathbb RP^2,y)$ is not the zero map. Note that there is no reason why $g(y') =y'$ for $y' \ne y$, thus we cannot change the basepoint in the fundamental group. Also note that $\pi_1(\mathbb RP^2,y) = \mathbb Z_2$, thus we must have $g_* = id$. We shall, however, not use this fact.
You correctly argue that your lift $h$ has the property that for all $x$ either $h(-x) = h(x)$ or $h(-x) = -h(x)$. But this does not a priori mean that either $h(-x) = h(x)$ for all $x$ or $h(-x) = -h(x)$ for all $x$. It could be that $h(-x_1) = h(x_1)$ for some $x_1$ and $h(-x_2) = -h(x_2)$ for some $x_2$. But if you think about it, you see that this is impossible: The sets $A = \{ x \in S^2 \mid h(-x) = h(x) \}$ and $B = \{ x \in S^2 \mid h(-x) = - h(x) \}$ are closed disjoint sets with $A \cup B = S^2$, thus either $A = X$ or $B = X$ since $S^2$ is connected.
Assume that $A = X$. Consider an element $u \in \pi_1(\mathbb RP^2,y)$ such that $g_*(u) \ne 0$. Represent $u$ by a loop $\gamma$ based at $y$. This loop lifts to a path $\gamma'$ in $S^2$ such that $\gamma'(0) = x \in p^{-1}(y)$. Since $\gamma'(1) \in p^{-1}(y)$, we have either $\gamma'(1) = x$ or $\gamma'(1) = -x$. But $h\gamma'$ is a path in $S^2$ from $h(\gamma'(0)) = h(x) = x'$ to $h(\gamma'(1)) = h(\pm x) = h(x) = x'$. Thus it is a loop at $x'$. Since $S^2$ is simply connected, we have $[h\gamma'] = 0 \in \pi_1(S^2,x')$. But $p(x') = ph(x) = gp(x) = g(y) = y$, thus $p_* : \pi_1(S^2,x') \to \pi_1(\mathbb RP^2,y)$ has the property that $0 = p_*([h\gamma']) = [ph\gamma'] = [gp\gamma'] = [g\gamma] =g_*([\gamma]) = g_*(u)$. This is a contradiction. Therefore $B = X$.