Show a connected subset of $\mathbb R^2$

Question

How can I show $\{(x_1,x_2)\in \mathbb R^2 :x_1^2+x_2^2=1\}$ is a connected subset of $\mathbb R^2$

MY SILLY ATTEMPTS INCOMING:

Call the function set above E. Say I assume $f(E)$ is not connected in $\mathbb R^2$. Can I say there are open sets $V_1$ and $V_2$ containing the function, that the intersection of the two open sets is empty, and that the union of them is nonempty

Then I could say that the inverse of the open sets, say $U_1=f^-1(V_1)$, etc, are open sets in $\mathbb R^2$ that separate $E$, and therefore $E$ is not connected...

I'm not sure how to do this, I was hoping to contradict myself up there haha

Answer 1

Hint: The set that you have mentioned is just $S^1$. The image of a connected set under a continuous map is connected . Consider the map $f:[0,2\pi]\to \Bbb{R^2}$ given by $f(x)=(\cos(x),\sin(x))$ . What can you say about it now?