I need to proof that $S^{n-1}:=\{x\in\mathbb{R}^{n}\, :\, ||x||=1\}$, $n>1$ is path connected.
So, for all $x,y\in S^{n-1}$, I need to show a function $f:[a,b]\rightarrow S^{n-1}$ such that $f$ is continuous, $f(a)=x$ and $f(b)=y$.
My proof: Let $x,y\in S^{n-1}$, and write it in polar coordinates, i.e., $$x=(1,\theta_{1},\phi_{1} )\quad\textrm{and}\quad y=(1,\theta_{2},\phi_{2}).$$
So, define a function $f:[0,1]\rightarrow S^{n-1}$ by $$f(t)=\left(1,\theta_{2}t+(1-t)\theta_{1},\phi_{2}t+(1-t)\phi_{1}\right)$$
So, $f$ is continuous, $f([0,1])\subset S^{n-1}$, $f(0)=x$ and $f(1)=y$.
Is my proof correct?
Your idea works for $n=2$. One way to extend to higher dimensions is to show first that antipodal points can be connected by a path. Then for any pair of non-antipodal points $x,y\in S^{n-1}$, let $\alpha(t)=(1-t)x+ty$ be the straight line connecting $x$ and $y$, and let $\tilde\alpha(t)=\frac{\alpha(t)}{\|\alpha(t)\|}$. Then $\tilde\alpha:[0,1]\to S^{n-1}$ is a path connecting $x$ and $y$.