Let $f: D \to \Omega$ be unif. cont. and $D \subseteq X$ is dense. Show that f has a unif. cont. extension in X. Can anyone verify this proof for me ? Thanks!
Suppose that $\Omega$ is a complete metric space and that $f: (D,d) \to (\Omega,p)$ is uniformly continuous, where $D$ is dense in $(X,d)$. Show that there is a uniformly continuous function $g: X \to \Omega$ with $g(x) = f(x)$, $\forall x \in D$.
$\\ \\ \textbf{Proof:} \\ \\$
By definition, a set D is dense if $\overline{D} = X$.
   Let $x \in \overline{D} = X$, then by Proposition
$\forall \epsilon > 0, B(x,\epsilon)\cap D \neq \emptyset$
Consider the sequence $x_n \in B(x,\frac{1}{n})\cap D$. This sequence is contained in $D$ and non-empty because $D$ is dense in $X$. Also this sequence is Cauchy, because $d(x_n,x_m) < \frac{1}{n} \to 0$. Since f is uniformly continuous $\{f(x_n)\}$ is also Cauchy in $\Omega$. By Continuity we have that
${x_n} \to x \Rightarrow f(x_n) \to f(x) = y_x$
By the completeness of $\Omega$, we have that $y_x \in \Omega$. Given $x \in X$, define $g: X \to \Omega$ as $g(x) = y_x$. By uniqueness of limits, this function is well-defined. For $x \in D$, define $g|_D = f$ and consider the sequence $x_n = x$ for all $n$. Clearly, $x_n \to x$ and hence
$g(x) = \lim_{n\to\infty}f(x_n) = \lim_{n\to\infty}f(x) = f(x), \forall x \in D.$
Now we must show that $g$ is uniformly continuous.
   Let $\epsilon > 0$, then $\exists \delta > 0$ such that
$\forall x,y \in D, d(x,y) < \delta \Rightarrow p(f(x),f(y)) < \epsilon.$
because $f$ is uniformly continuous on $D$. Now pick $x, x^0 \in X$ such that $d(x,x^0) < \frac{\delta}{3}$. Using the previous construction we can find $\{x_n\}$ and $\{x_n^0\}$ in $D$ such that $x_n \to x$ and $x_n^0 \to x^0$.
By the convergence of the sequences $\{x_n\}$ and $\{x_n^0\}$ we can pick $N_1 \in \mathbb{N}$ such that for all $n \geq N_1$ $d(x_n,x) < \frac{\delta}{3}$ and $d(x_n^0,x^0) < \frac{\delta}{3}$ . Then by triangle inequality
$d(x_n,x_n^0) \leq d(x_n,x) + d(x,x_n^0) + d(x_n^0,x^0) < \frac{\delta}{3} + \frac{\delta}{3} + \frac{\delta}{3} = \delta.$
By the definition of $g$
$f(x_n) \to g(x) \in X \mbox{ and } f(x_n^0) \to g(x^0) \in X.$
Then there exists $N_2 \in \mathbb{N}$ such that for all $n\geq N_2$
$p(f(x_n),g(x)) < \epsilon \mbox{ and } p(f(x_n^0),g(x^0)) < \epsilon.$  
Now fix $N_0 > \mbox{max}\{N_1,N_2\}$, then applying triangle inequality twice
$\begin{align*}
   p(g(x),g(x_0)) &\leq p(g(x),g(x_0^n)) + p(g(x_0^n),g(x)) \\
   &\leq p(g(x),g(x_0^n)) + p(g(x_0^n),g(x_n)) + p(g(x_n),g(x)) \\
   &= p(g(x),f(x_0^n)) + p(f(x_0^n),f(x_n)) + p(f(x_n),g(x)) \\
   & < \epsilon + \epsilon + \epsilon = 3\epsilon.
   \end{align*}$
Hence, $g$ is uniformly continuous.
 A: You put $f(x)=y_x$ and defined $g(x)=f(x)$. Unfortunately, we didn't define $f$ at general $x\in X$, so this definition doesn't make sense.
Probably you intened to define $g(x)$ as the limit of $f(x_n)$, which can be defined by completeness. However this $g(x)$ is not well-defined, since $\lim_{n\to\infty} f(x_n)$ may vary when we take different sequence $x_n\to x$. So you need to prove the well-definedness of $g$, and it can be verified using uniform continuity of $f$. It's not necessary to restrict the distance of $x_n$ and $x$ to be less than $1/n$.
The rest part of your proof is correct. I just would like to point out that no additional condition is needed to $x_n$ and $x_n ^0$, except convergence to corresponding points.

The well-definedness of $g$ can be established as follows: Take $\delta>0$ to satisfty $p(f(x),f(y))<\epsilon$ if $d(x,y)<\delta$. Now assume that $x_n\to x$ and $y_n\to x$. Then, by the uniform continuity of $f$, there is a natural number $N$ such that $d(x_n ,x)<\delta /2$ and $d(y_n ,x)<\delta /2$ for every $n>N$.
To sum up, for every $\epsilon>0$ there is a $N$ such that $p(f(x_n),f(y_n))<\epsilon$ for every $n>N$. Now put $c=\lim_{n\to\infty} f(x_n)$ and take integers $M_1 ,M_2$ such that
$$n>M_1 \Rightarrow p(f(x_n),c)<\epsilon/2$$
$$n>M_2 \Rightarrow p(f(x_n),f(y_n))<\epsilon/2$$
Then for every $n>\max (M_1,M_2)$, $p(f(y_n),c)<\epsilon$. Hence $f(y_n)\to c$.
