I seem to be stuck on. I think I figured out how to show $x^2 + y^2 = 1$ is connected I believe. If $X$ is connected and $f:X \to Y$ is continuous, then $f(x)$ is connected. By setting $X=[0, 2\Pi]$, and $f$ as the function $(\cos x, \sin x)$, then $f(X)$ is the image of the function in that domain so it must be connected. However, I can't seem to figure it out with 3 variables.
Edit: $x^2 + y^2 + z^2 = 1$
A similar argument works. The function $$f:[0,2\pi]\times[0,2\pi]\to S^2$$ given by $$(u,v)\mapsto (\cos u \cos v,\cos u \sin v, \sin u)$$ is continuous onto $S^2$ and since squares are connected, the sphere is, too.
You might want to restrict the domain, but for the purposes of your argument it works this way.