Let $\Omega\subset \mathbb{R}^n$ and $f:\Omega\rightarrow \mathbb{R}^m$ uniformly continuous. Prove that exists $\overline{f}:\overline{\Omega}\rightarrow \mathbb{R}^m$ uniformly continuous in$\overline{\Omega}$ such that $\overline{f}(x)=f(x)$, for all $x\in \Omega$.
Attempt: well, first I put my eyes on a point $x_0$ in the boundary of Omega, then, $x_0$ is an accumulation point, which implies that there exists a sequence {$x_n$} in Omega that converges to the point $x_0$. So I guess I need to manipulate the images of $f$ under the terms of the sequence and see that the limit of the sequence of the images exists, but I'm so confused, and I don’t know where I need to start