Let U be open in $\Bbb{R}^n$ such that $U\neq \emptyset$ . Prove that if U is complete, then $U=\Bbb{R}^n$

Question

I have this interesting problem.

Let U be open in $\Bbb{R}^n$ such that $U\neq \emptyset$ . Prove that if U is complete, then $U=\Bbb{R}^n$.

According to my lecturer, this could be a one-line proof but we (students) couldn't even see any solution coming. Please, is there any way of solving this?

I know that if U is open, then $\forall x\in U,\;\;\exists\;r>0\;:\;B(x,r)\subset U.$

If $U$ is complete, then every Cauchy sequence in $U$ has a convergent subsequence.

Thanks for your time and help!

Answer 1

If $U$ is complete then $U$ is closed, so $U$ is clopen, but $U$ is non-empty and $\Bbb R^n$ is connected, so $U$ is the whole space.