Let $f(x)$ and $g(x)$ be continuous mappings from topological space $X$ to $S^n$, such that $f(x)$ is different from $-g(x)$ for all $x$. Prove that $f$ and $g$ are homotopic.
So, my idea was to start off with the definition of homotopy: Let $F: X \times I\to S^n$ $F(x,0)=f(x)$, $F(x,1)=g(x)$, $F(x,t)=tg(x)+(1-t)f(x)$. But how to prove that $F(x,t)\in S^n$?
Defining $F(x,t)=tg(x)+(1-t)f(x)$ is a nice try. However, $tg(x)+(1-t)f(x)\not\in S^n$ in general, and we only know that $tg(x)+(1-t)f(x)\in\mathbb{R}^{n+1}$. Therefore, we have to divide it by its norm. That is, we define $$F(x,t)=\frac{tg(x)+(1-t)f(x)}{\|tg(x)+(1-t)f(x)\|}.\tag{0}$$
Now, we have to worry if $\|tg(x)+(1-t)f(x)\|=0$, which would make the map not well-defined. That's where the assumption comes in. If $\|tg(x)+(1-t)f(x)\|=0$ for some $(x,t)\in X\times I$, then we have $tg(x)+(1-t)f(x)=0$, which gives $$tg(x)=-(1-t)f(x).\tag{1}$$ Taking the norm of both sides, we get $$t\|g(x)\|=(1-t)\|f(x)\|\tag{2}$$ since $t, 1-t\geq 0$. Using the facts that $\|g(x)\|=\|f(x)\|=1$ (since $g, f:X\to S^n$), we obtain from $(2)$ that $t=1-t$. Putting this back to $(1)$, we obtain $g(x)=-f(x)$, which contradicts to the assumption.
Therefore, $F$ defined in $(0)$ is well-defined, and $F:X\times I\to S^n$. And $F$ is the required homotopy between $f$ and $g$, which I invite you to check.