Proving elementary statements about a metric space I am trying to solve the following exercise in Kolmogorov's real analysis textbook.

Given a metric space $R$, let $A$ be a subset of $R$ and $x$ a point of $R$. Then the number 
  \begin{equation}
\rho(A,x) = \inf_{a \in A} \rho(a,x) 
\end{equation}
  is called the distance between $A$ and $x$. Prove that 
(a) $x \in A$ implies $\rho(A,x) = 0$, but not conversely;
(b) $\rho(A,x)$ is a continuous function of $x$ (for fixed $A$); 
(c) $\rho(A,x) = 0$ if and only if $x$ is a contact point of $A$; 
(d) $[A] = A \cup M$, where $M$ is the set of all points $x$ such that $\rho(A,x) = 0$. 

Here is my attempt at a solution, with notes on where I am currently stuck. 

(a) Let $x \in A$. Then, we have: 
  \begin{align*}
\rho(A,x) = \inf_{a \in A} \rho(a,x) = \rho(x,x) = 0,
\end{align*}
  since, by the properties of the distance function, the distance between a point and itself is $0$. 

I cannot figure out how to prove the converse. I know that we want to refute the statement that for any $x$, $\rho(A,x) = 0$ implies that $x \in A$, so we need to find an $x$ such that $\rho(A,x) = 0$ but $x \not \in A$. If $x \not \in A$, it must be the case that $x \in R \setminus A$. Since $\rho$ is positive definite, though, $\rho(a,x) = 0$ if and only if $x = a$. The key then must be in the infimum, which would suggest that we can pick an $x$ such that our distance gets arbitrarily close to $0$, but does not actually get to $0$. Since I can't specify $x$ as a sequence of values, I am stuck. 

b) To show that $\rho(A,x)$ is continuous on $R$, we need to prove that: for any $x_0 \in R$ and $\epsilon > 0$, there exists a $\delta > 0$ such that for any $x \in R$, if $|x-y| < \delta$, $|\rho(A,x) - \rho(A,y)| < \epsilon$. 

I cannot figure out how to work in the assumption that $|x-y| < \delta$, as we can't map this statement onto a statement about the infimum, which is what we currently have. I can try to manipulate the statement by adding and subtracting terms and using the triangle inequality, but haven't been able to get anywhere.

We suppose $x$ is a contact point of $A$. By definition, every neighborhood of $x$ contains a point of $A$. That is, every neighborhood of $x$ contains a point with distance $0$ from some point of $A$. 

I am not completely certain on this proof, as this doesn't necessarily seem to imply that $x \in A$. I considered an argument by contradiction, which would involve finding supposing that the infimum is larger and then producing a smaller epsilon. I can't figure out how to appropriately formalize this argument. 

Alternatively, we suppose $\rho(A,x) = 0$. So 
  \begin{equation*}
\rho(A,x) = \inf_{a \in A} \rho(a,X) = 0.
\end{equation*}

I am not sure if I am allowed here to say that $x \in A$. If so, the proof is complete, so an element of $A$ is surely a contact point of $A$, but it not, then there is a similar issue as before as it relates to approaching, but not hitting $0$. 

(d) We denote the closure of $A$ by $[A]$. We will demonstrate set equality by dual containment. 
Let $x \in A \cup M$, so $x \in A$ or $x \in M$ or both. If $x \in A$, then $x$ is a contact point of $A$, so $x \in [A]$. If $x \in M$, then $\rho(A,x) = 0$. By part (c), $x$ is a contact point of $c$, so $x \in [A]$. If $x$ is in both, $x \in [A]$, so $A \cup M \subset [A]$. 

The proof for this half of the containment seems straightforward. I am less certain on the other half. 

Let $x \in [A]$. Then $x$ is a contact point of $A$, so every neighborhood of $x$ contains at least one point of $A$. 

Surely a contact point of $A$ is an element of $A$ and a point such that $\rho(A,x) = 0$ is also a contact point. It seems, though, that I'm still establishing the opposite implication. 
Any help or feedback on this would be greatly appreciated. 
 A: For the converse of item (a), consider $\mathbb{R} $ with the usual metric and the set $A = \left\lbrace \dfrac{1}{n}, n \in \mathbb{N} \right\rbrace \subset \mathbb{R}$.
$$\rho(0, A) = \text{inf} \left\lbrace \left|0 - \dfrac{1}{n}\right| , n \in \mathbb{N} \right\rbrace$$
$$\rho(0, A) = \text{inf}\left \lbrace \dfrac{1}{n} , n \in \mathbb{N} \right\rbrace = 0$$
Thus, $\rho(0, A) = 0$ and $0 \not\in A$.
A: $b)$.Let $x,y \in X$
Then $\forall z \in A$ we have $$d(x,A) \leq d(x,z)  \leq d(x,y)+d(y,z) \Longrightarrow d(x,A) \leq d(x,y)+d(y,A) \Longrightarrow d(x,A)-d(y,A) \leq d(x,y)$$
Similarly you can prove that $d(y,A)-d(x,A) \leq d(x,y)$
So $|d(x,A)-d(y,A)| \leq d(x,y)$ (you can finish it form here...)
$------------------------------------------$
$c).$ $(\Longrightarrow)$ Since $\inf_{a \in A}d(x,a)=0$ then  we can find $a_n \in A$  such that $d(x,a_n) \leq \frac{1}{n} \Longrightarrow a_n \to ^d x \Longrightarrow x$ is a contanct point of $A$
$(\Longleftarrow)$Let $x$ a contanct point of $A$. Then exists $a_n \in A$ such that $a_n \to^d x$. In other words $d(a_n,x) \to 0$.
But $\forall n \in \Bbb{N}$ we have that $d(x,A) \leq d(x,a_n) \to 0 \Longrightarrow d(x,A)=0$
$------------------------------------------$
$d).$ Its a consequence of $c$ and the definition of the closure of a set.
