If $f\colon S^n \longrightarrow S^m$ continuous, $m\geq n$.

You can prove that $f^*$ must be an isomorphism by using that $f_{\#}$ must be an isomorphism on fundamental groups and then using both Hurewicz and the Universal Coefficients Theorem.

To prove that $f_{\#}$ is an isomorphism on fundamental groups, you indeed need that $f(x)=-f(-x)$. You pick a point $p$ and a path $\gamma$ taking $p$ to $-p$ on the sphere $S^n$. The path $f \circ \gamma$ does the same thing (takes a point $p'$ to $-p'$) by the property that $f(x)=-f(-x)$, and thus the map induced in projective spaces induces an isomorphism in fundamental groups (it is taking a non-trivial loop to a non-trivial loop, and the fundamental group is $\mathbb{Z}_2$).

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However, there is a more elementary approach (which can be seen in Bredon):

Given a two-sheeted covering $p:X \to Y$ (in this case, $p_n: S^n \to \mathbb{R}P^n$), there is a long exact sequence $$\cdots \to H_p(Y;\mathbb{Z}_2) \stackrel{t_*}{\to} H_p(X,\mathbb{Z}_2) \stackrel{p_*}\to H_p(Y;\mathbb{Z}_2) \stackrel{\partial_*}\to H_{p-1}(Y;\mathbb{Z}_2) \to \cdots, $$ which is obtained from the short exact sequence of complexes $$0 \to \Delta_p(Y;\mathbb{Z}_2) \stackrel{t}{\to} \Delta_p(X,\mathbb{Z}_2) \stackrel{p_{\Delta}}\to \Delta_p(Y;\mathbb{Z}_2) \to 0 ,$$ where the only map which needs explaining is $t$: this is the map taking $\sigma$ to $\sigma +a \circ \sigma,$ where $a$ is the only deck transformation which is not the identity. You can check that this sequence is indeed exact.

For your case of $S^n$ and $\mathbb{R}P^n$, we have $$0 \to H_m(\mathbb{R}P^m;\mathbb{Z}_2) \to H_m(S^m,\mathbb{Z}_2) \to H_m(\mathbb{R}P^m;\mathbb{Z}_2) \to H_{m-1}(\mathbb{R}P^m;\mathbb{Z}_2) \to \cdots, $$ $$\cdots \to H_1(\mathbb{R}P^m;\mathbb{Z}_2) \to H_0(\mathbb{R}P^m;\mathbb{Z}_2) \to H_0(S^m,\mathbb{Z}_2) \to H_0(\mathbb{R}P^m;\mathbb{Z}_2) \to 0. $$ We have that the above sequence is $$0 \to H_m(\mathbb{R}P^m;\mathbb{Z}_2) \stackrel{\simeq}{\to} H_m(S^m,\mathbb{Z}_2) \stackrel{0}\to H_m(\mathbb{R}P^m;\mathbb{Z}_2) \stackrel{\simeq}\to H_{m-1}(\mathbb{R}P^m;\mathbb{Z}_2) \to \cdots, $$ $$\cdots \stackrel{0}\to H_1(\mathbb{R}P^m;\mathbb{Z}_2) \stackrel{\simeq}{\to} H_0(\mathbb{R}P^m;\mathbb{Z}_2) \stackrel{0}\to H_0(S^m,\mathbb{Z}_2) \stackrel{\simeq}\to H_0(\mathbb{R}P^m;\mathbb{Z}_2) \to 0. $$ This is trivial if you know the homology of $\mathbb{R}P^n$ and $S^n$ (the first map is injective, but since it is a map from $\mathbb{Z}_2$ to $\mathbb{Z}_2$ it must be an isomorphism etc), but could also be proven without resorting to knowing the explicit homology of $\mathbb{R}P^n$, if you go down to using the definitions of the maps.

This tells us that each $\partial_*$ is an isomorphism.

Let's now assume that $n>m$. Naturality of $\partial_*$ tells us that the following diagram commutes for all $i$ $$\begin{array}{ccccccccc} H_i(\mathbb{R}P^n;\mathbb{Z}_2) & \xrightarrow{\partial_*} & H_{i-1}(\mathbb{R}P^n;\mathbb{Z}_2) \\ \downarrow{f_*} & & \downarrow{f_*} \\ H_{i}(\mathbb{R}P^m;\mathbb{Z}_2) & \xrightarrow{\partial_*} & H_{i-1}(\mathbb{R}P^m;\mathbb{Z}_2).\end{array}$$ By induction and since $\partial_*$ is an isomorphism up to $m$ (both above and below, since $n>m$) by what we've seen above and also since $f_*$ is obviously a isomorphism when $i-1=0$, it follows that $f_*$ is an isomorphism up to when $i=m$.

But this contradicts the following commutative diagram $$\begin{array}{ccccccccc} H_m(\mathbb{R}P^n;\mathbb{Z}_2) & \xrightarrow{t_*} & H_{m}(S^n;\mathbb{Z}_2) \\ \downarrow{f_*} & & \downarrow{f_*} \\ H_{m}(\mathbb{R}P^m;\mathbb{Z}_2) & \xrightarrow{t_*} & H_{m}(S^m;\mathbb{Z}_2),\end{array}$$ since going from top left to down right is a composition of isomorphisms of non-trivial groups, whereas the top right is $0$.