Extension of a bounded and uniformly continuous function I read in a book the following assertion:

Let $\Omega\subset\mathbb{R^n}$ be a open set, then any bounded and uniformly continuous function in $\Omega$ has a unique bounded continuous extension to $\overline{\Omega}$.

I can't see this readly, first I thought was a simple consequence of the Tietze extension theorem, but I didn't have any progress. Someone could help me?
Thanks! 

For any $x \in \overline{\Omega}$, find a sequence $(x_n) \in \Omega$ such that $x_n \to x$.


*

*Since $(x_n)$ is Cauchy, and $f$ is uniformly continuous, show that $(f(x_n))$ is Cauchy.

*Define $\overline{f}(x) = \lim_{n\to \infty} f(x_n)$

*Show that this definition is independent of the choice of the sequence $(x_n)$.


Now $\overline{f}$ is your extension (it coincides with $f$ because of #3)
Edit: Uniqueness holds because any continuous extension of $f$ must satisfy #2 : ie. if $g$ is another continuous extension of $f$, then for any $x_n$ as above
$$
g(x) = \lim_{n\to \infty} g(x_n) = \lim_{n\to \infty} f(x_n) = \overline{f}(x)
$$
As for boundedness, it again follows from #2 : If $|f(y)| \leq M$ for all $y \in \Omega$, then $|\overline{f}(x)| = \lim_{x_n\to x}|f(x_n)| \leq M$ as well.
