Prove that the unit circle is path-connected?

Your function $f$ can be used to show that the unit circle is in fact path-connected: pick $\mathbf{x}\neq\mathbf{y}$ in the unit circle. Since your $f$ is onto, there exist $a,b\in [0,2\pi]$ such that $f(a)=\mathbf{x}$, $f(b)=\mathbf{y}$. Assume, without loss of generality, that $a\lt b$. Then consider a map $g\colon[0,1]\to[a,b]$ given by $g(t) = a+t(b-a)$. Then consider the function $f\circ g\colon [0,1]\to \mathbf{C}$.

Hummm, I'm not a native English speaker, so I am a bit lost concerning the mathematical vocabulary, but I'll try my best.

For proving that the unit circle is connected, you could also say that "the only subsets of the unit circle which are both open and closed are the full circle and the empty set". Which is obvious when you think about it.

For the path-connection, I'm taking the definition : a set is path connnected if, for any pair of points you can make a path (continuous function on$[0,1]$) that joins them.

Basically, for the two points $a_1$ and $b_1$ of your comment, $f(x)=(r\cos(t*(x_2-x_1)+x_1),r\sin(t*(x_2-x_1)+x_1))$ is a path that links these two points ($f(0) = a_1 \text{ and } f(1)=b_1$).

P.S : This is my first post, so where can I find a faq about how to write mathematical expressions correctly?