I had a lot of help on this question in chat from users Srivatsan and t.b. the other day. I tried my best to write up what was said as an answer here.
Notice that the sets $\overline{f(V_n(p))}\supseteq\overline{f(V_{n+1}(p))}\supseteq\cdots$ form a nested sequence of closed sets. Moreover, let $\epsilon>0$ be given. Since $f$ is uniformly continuous, there exists $\delta>0$ such that $d(p,q) < \delta$ implies $d(f(p),f(q))<\epsilon$ for $p,q\in E$. Taking $n$ large enough so that $\frac{2}{n}<\delta$, then for $q,r\in V_n(p)$,
$$
d(q,r)<d(q,p)+d(p,r)<\frac{2}{n}<\delta
$$
so $d(f(q),f(r))<\epsilon$. Thus $f(V_n(p))$ is bounded in $\mathbb{R}$, so $\overline{f(V_n(p))}$ is bounded as well. Hence for large enough $n$ the sets form a compact nested sequence. Since $V_n(p)$ has diameter at most $\frac{2}{n}$, taking $n>2/\delta$ would imply
$$
\operatorname{diam } f(V_n(p))=\operatorname{diam }\overline{f(V_n(p))}<\epsilon
$$
So $\lim_{n\to\infty}\operatorname{diam }\overline{f(V_n(p))}=0$, and thus their intersection consists of a single point. Also, since $\operatorname{diam }f(V_n(p))\to 0$ as $n\to\infty$, and so by choosing points arbitrarily close to $p$, their images under $g$ are arbitrarily close to $g(p)$. (To be more explicit, letting $\delta$ be small enough such that for $x,y\in E$, then $d(x,y)<2\delta$ implies $d(f(x),f(y))<\epsilon/3$, choose $n$ large enough that $\frac{1}{n}<\delta$, and thus for any $x,y\in V_n(p)$, $d(x,y)<2/n<2\delta$, so $\operatorname{diam }f(V_n(p))<2\epsilon/3$, so $d(f(x),g(p))<2\epsilon/3$. Note also that this can be done for any $p$.)
I contend that $g$ is uniformly continuous. Let $\epsilon>0$ be given. Since $f$ is uniformly continuous, there exists $\delta>0$ such that $d(r,s)<\delta$ implies $d(f(r),f(s))<\epsilon/3$. Now let $p,q\in X$ be any points such that $d(p,q)<\delta/3$. By the above reasoning, choose $n$ large enough so that $n>\frac{3}{\delta}$, and both $d(g(r),g(p))<\epsilon/3$ and $d(g(s),g(q))<\epsilon/3$ for $r\in V_n(p)$ and $s\in V_n(q)$. Also,
$$
d(r,s)<d(r,p)+d(p,q)+d(q,s)<\delta
$$
so $d(f(r),f(s))=d(g(r),g(s))<\epsilon/3$. By the triangle inequality, $d(g(p),g(q))<\epsilon$, so $g$ is uniformly continuous, and thus continuous on $X$, and of course $g|_E=f$.
(Your proof above should explicitly show that $g$ is independent of the sequence used to define it. This is the key point of the proof.)
Let $\epsilon>0$, then you have some $\delta>0$ such that if $d(x,y) < \delta$, then $d(f(x),f(y)) < {1 \over 2}\epsilon$.
Pick $x,y \in \overline{A}$ such that $d(x,y) < \delta$, and let $x_n,y_n$ be sequences in $A$ such that
$x_n \to x,y_n \to y$. By construction above, $g(x) = \lim_n f(x_n)$ and similarly for $g(y)$.
For sufficiently large $n$, we have $d(x_n,y_n) < \delta$, and so $d(f(x_n),f(y_n)) < {1 \over 2}\epsilon$.
Taking limits we have $d(g(x),g(y)) \le {1 \over 2}\epsilon < \epsilon$.
Best Answer
You may want to see the answers for this question, which answer yours, Extending a function by continuity from a dense subset of a space.
I built the proof myself based on Srivatsan's answer for that question. If anybody still needs it, here it goes:
Theorem
If $X$ and $Y$ are metric spaces and $f:S \to Y$ is uniformly continuous with $S$ dense in $X$, and $Y$ is complete, then there exists a unique continuous extension of $f$ in $\overline{S}$ which by the way is uniformly continuous.
Proof
Let $d$ and $D$ be the metrics of $X$ and $Y$ respectively.
Let $g:\overline{S} \to Y$ be given by $g(a) = \lim f(x_n)$, where $(x_n)$ is any sequence of points in $S$, with $x_n \to a$.
$g$ is well defined:
$\lim f(x_n)$ exists:
Let $\varepsilon > 0$. Because of the uniform continuity of $f$, there exists $\delta>0$ such that for every $a,b \in S$, if $d(a,b) < \delta$, then $D(f(a),f(b)) < \varepsilon$.
Since $x_n \to a$, $(x_n)$ is Cauchy, there exists $N \in \mathbb{Z}^{+}$ such that if $n,m \geq N$, $d(x_n,x_m)<\delta$.
Hence, if $n,m \geq N$, $D(f(x_n),f(x_m))<\varepsilon$. Then $(f(x_n))$ is Cauchy, and since $Y$ is complete, $\lim f(x_n)$ exists.
If $x_n \to a$ and $y_n \to a$ then $\lim f(x_n) = \lim f(y_n)$:
Let $(z_n) = (x_1,y_1,x_2,y_2,...)$. If $\varepsilon>0$, there exists $N \in \mathbb{Z}^{+}$ with $d(x_n,a) < \varepsilon$ and $d(y_n,a) < \varepsilon$ for each $n \geq N$.
Consequently, if $n \geq 2N$, then $n/2,(n+1)/2 \geq N$ and so, if $n$ is even, $d(z_n,a) = d(y_{n/2},a) < \varepsilon$, and if $n$ is odd, $d(z_n,a) = d(y_{(n+1)/2},a) < \varepsilon$. Therefore $z_n \to a$.
So, $\lim f(z_n)$ exists and since $(f(x_n))$ and $(f(y_n))$ are subsequences of $(f(z_n))$, $\lim f(x_n) = \lim f(z_n) = \lim f(y_n)$.
$g$ is an extension of $f$:
$g$ is uniformly continuous:
Let $\varepsilon > 0$. Since $f$ is uniformly continuous, there exists $\delta > 0$ such that $D(f(a),f(b))<\varepsilon/3$ for every $a,b \in S$ with $d(a,b)<\delta$.
Let $a,b \in \overline{S}$ with $d(a,b)<\delta/3$.
There exist sequences in $S$, $(x_n)$ and $(y_n)$ with $x_n \to a$ and $y_n \to b$. Since $x_n \to a$ and $y_n \to b$, there exists $N_1 \in \mathbb{Z}^{+}$ with $d(x_n,a)<\delta/3$ and $d(y_n,b)<\delta/3$ for every $n\geq N_1$.
If $n \geq N_1$, $d(x_n,y_n) \leq d(x_n,a) + d(a,b) + d(b,y_n) < \delta$ and so, $D(f(x_n),f(y_n)) < \varepsilon/3$.
Also, since $f(x_n) \to g(a)$ and $f(y_n) \to g(b)$, there exists $N_2 \in \mathbb{Z}^{+}$ with $D(f(x_n),g(a))<\varepsilon/3$ and $D(f(y_n),g(b))<\varepsilon/3$ for every $n\geq N_2$.
Then, if $N=max\{N_1,N_2\}$, $D(g(a),g(b)) \leq D(g(a),f(x_N)) + D(f(x_N),f(y_N)) + D(f(y_N),g(b)) < \varepsilon.$
$g$ is unique: