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 $g|_E=f$.
Best Answer
In short: Your solution is correct.
You also already know that the optimal radius is less than that and we can go on step in that direction: Let $D=\operatorname{diam}(A)$. As also $\operatorname{diam}(\overline A)=D$, we may assume wlog that $A$ is compact, so there are points $a,b\in A$ with $d(a,b)=D$. Then $$A\subseteq\overline{B(a,D)}\cap \overline{B(b,D)} $$ The lens shape on the right can be covered from its center $c=\frac{a+b}2$ with radius $r$ given by Pythagoras, i.e., $(D/2)^2+r^2=D^2$.$^1$ In other words, this already improves our result to $$ A\subset \overline{B\bigl(\tfrac{a+b}2,\tfrac{\sqrt3}2D\bigr)}$$
$^1$ That is: A quick computation shows that $(x-a)^2\le D^2 \land (x-b)^2\le D^2$ implies $(x-c)^2\le \frac34 D^2$