I want to show that $\pi_n(\mathbb{R}P^m)$ is trivial for $1<n<m$.
I've managed to do that via lifting and Sard Theorem. Indeed, let $f:S^n \to \mathbb{R}P^m$. Since $S^n$ is simply connected, it lifts to a map $\widetilde{f}:S^n \to S^m$. It suffices then to show that such lift is homotopically trivial. We can approximate homotopically such a map by a smooth function. By Sard Theorem, such approximation has a regular value. Since $m>n$, it follows that there is a point which is not on the image. Therefore, our lift factors through a map $\widetilde{f}':S^n \to \mathbb{R}^m$. Since $\mathbb{R}^m$ is contractible, the result follows.
Now, my issue is the following. This kind of approach is often used, for example, to prove that $\pi_1(S^n)=0$ for $n>1$. However, Bredon has a nice workaround which does not used Sard theorem in this case. He uses uniform continuity to guarantee the fact that you can make your loop be homotopically equivalent to concatenations of arcs of great circles. My question is:
What would be a way to prove the fact in the title without using Sard's theorem/smooth approximation?