closure and metric spaces proof Let (M, d) be a metric space and K ⊂ M. For x ∈ M
Define: $ d(x, K) = inf{d(x, y); y ∈ K} $
show that $d(x,K) = 0$ if and only if x is an element of closure(K) 
%% it was K bar in their notation.
I'd appreciate it if you tell me what you think of my proof so far:
case 1: let x be an element of K. Then d(x, K) = 0 because 
inf{d(x,y); y ∈ K} = d(x,x) = 0. The distance between a point and itself is zero.
case 2: let x  ∈ K' (the set of all limit points of K).
this is a bit trickier and I'm not sure how to word it.
Since x is a limit point of K, then there are an infinite number of points around x, which are still in K, which do not equal x.
In this set, we can get arbitrarily close to x.
{x + d, x + 2d, x + 3d...}
the infimum of the set of distances between each point in this set and x is:
{x + d - x, x + 2d - x, x + 3d - x}
= {d, 2d, 3d,...} for arbitrarily small d.
We will never get x, so we will never get a distance of zero, but we will get every other positive number: since the definition of a limit point is that every neighborhood contains a point in K which is not x.
Therefore, the greatest upper bound of this set is the greatest lower bound on the set of all positive real numbers, which is 0.
therefore in case 1 and case 2, the infimum of the desired set is 0.
Any input or clarity would be great! thanks
so if x is an element of K', then inf{d(x, y); y ∈ K = 0
 A: (1) For the forward direction, suppose $d(x,K) = 0$. Then there exists $y \in K$ with $d(x,y)=0$ and so we have x = y (by positive-definiteness of the metric) i.e $x \in K \subset \overline{K}$. 
(2) For the converse, if we suppose $x \in \overline{K}$ then for all $\epsilon>0$ we have $B(x, \epsilon) \cap K \not = \emptyset$. If $d(x,K)>0$ say $r$ then $B(x,\delta) \cap K = \emptyset$ (where $\delta<r$) which is a contradiction and so $d(x,K) = 0$. 
A: For case 2, you may formulate your argument as follows: Since you are working on a metric space, it is valid to use the term "open balls". Let $x$ be a limit point of $K$, then for every natural number $n$, there exists an open ball $B_n$ centred at $x$ of radius $\frac{1}{n}$ such that $(B_n\setminus\{x\})\cap K\neq\emptyset$. Then choose $y_n\in(B_n\setminus\{x\})\cap K$, we have $d(x,y_n)<\frac{1}{n}$. It follows that
$$d(x,K)\leq\frac{1}{n},\quad \forall n\in\mathbb{N}.$$
By letting $n\to\infty$, we have
$$d(x,K)\leq0.$$
As $d(x,K)$ is always non-negative, it follows that $d(x,K)=0.$
On the other hand, you still need to prove that $d(x,K)=0$ implies $x\in\overline{K}$. This is not terribly hard. Assume the contrary that $d(x,K)=0$, but $x\notin \overline{K}$. Note that $\overline{K}^c$ is open, then there exists $\epsilon>0$ such that the open ball $B_\epsilon$ which is centred at $x$ with radius $\epsilon$, is disjoint from $\overline{K}$. This means for any $y\in\overline{K}$, we have
$$d(x,y)\geq\epsilon>0,$$
which implies
$$d(x,K)\geq\epsilon>0,$$
thus we get a contradiction.
