Let $x_0$ be a point of $S^2$ such that it's not in the image of $\alpha(I)$, then $S^2$ is homomorphic to a plane. Consider $\gamma(s,t)=(1-t)\alpha(s)+t\beta(s)$ and $\beta(s)=(1-s)\alpha(0)+s\alpha(1)$ for all $t,s\in\,I$.
So,
$$\gamma(s,0)=\alpha(s)\;\;\gamma(0,t)=\alpha(0)=\beta(0)\;\;\gamma(s,1)=\beta(s)\;\;\gamma(1,t)=\beta(1)=\alpha(1)$$