continuity and closure questions - topology Let $(X,d)$ be a metric space. Let $U \subseteq (X,d)$. let $k \in (X,d)$.
Prove that if $U$ is fixed, $d(U,k)$ is a continuous function of $k$.
Prove that $\overline{U} = U \cup V$ where $V$ is the set of all points $k$ s.t. $d(U,k) = 0$
i.) Continuity states that $\forall \epsilon > 0, \exists \delta > 0$ s.t. if $d(x,y) < \delta$ then $d(f(x),f(y)) < \epsilon$. I wasn't sure how to apply this here as I don't know how to approach it when they say $U$ is fixed.
ii.) The way $V$ is defined, the points that satisfy this are the points on the "edge" of an open ball. So if one thinks of $B(2,3)$, then $k$ would be $2$ and $3$. $U$ would be all the interior points of $U$. Hence taking the union of these gives us the closure of $U$. 
These were my ideas. The help would be appreciated!

Here is a hint for i):
We have $f(y)=d(U,y)=\inf\{d(u,y)\mid u\in U\}$. Since for each $v\in U$
$$d(v,y)\le d(v,x)+d(x,y)$$
by the triangle equality, it follows that
$$d(U,y)\le d(v,y)+d(x,y)$$
for each $v\in U$, and thus
$$d(U,y)\le d(U,x)+d(x,y)$$
Can you do the rest?
For ii):
I assume the closure of a set $U$ is defined as the union $U\cup U'$, where $U'$ is the set of limit points of $U$, that is, $x\in U'$ if for each $\varepsilon>0$ there is a $u\in U$ such that $0<d(u,x)<\varepsilon$ (Let me know if you use a different definition of "closure").
Then you have to show two inclusions


*

*$U\cup V\subseteq U\cup U'$:
Show that each point in $V$ is either a point in $U$, or is a limit point of $U$. What does it mean that $\inf\{d(u,x)\mid u\in U\}=0$?

*$U\cup U'\subseteq U\cup V$:
Show that each limit point of $U$ has distance $d(U,x)=0$ from $U$.

