Sufficient Condition for the Existence of a Midpoint of a Chord We have an arbitrary metric space $X$ for which there exists a subset $A \subseteq X$ satisfying $A' \neq\varnothing$ (the set of limit points of $A$).  What do we need to assume about $X$ in order for the following to be true:
$$
    \forall a,b \in X \hspace{1mm} \exists c\in X \colon d_X(a,c) = d_X(b,c)
$$
where $d_X$ denotes the metric of $X$?
Certainly, this is not true for any metric space. Take, for a counterexample,
$X=[0,10] \setminus\{5\}$ (endowed with the absolute value function), $a=2$, and $b=8$. 
For context: I'm an undergrad who has never taken a course on topology.
Edit 1: Although it now appears to be a red herring, my original guess was connectedness.
Edit 2:
I am working on a problem (of which this is part) where I have already assumed the following about the metric $d_X$ of $X$:
$$
d_X(x,y) d_X(z,0) = d_X(zx, zy)
$$
$$
d_X(x,y) = d_X(x -y , 0)
$$
$$
d_X(x + y , 0) \leq
    d_X(x,0) + d_X(y,0)
$$
for any $x,y,z\in X$. Granted, this is somewhat coincidental, and the original question can be seen in a broader context. What originally motivated the question is my desire (in a proof I'm working on) to consider a neighborhood $N$ in $X$ which satisfies $a,b \in \partial(N)$. For concreteness, it oringinally seemed easiest to do so by considering a neighborhood $N_{d_X(c,a)}^X(c)$ where $c$ satisfies $d_X(a,c) = d_X(b,c)$. However, this requires that such a $c$ exists; hence, the original question.
Thank you.
 A: I've been Googling, but I can't find anything directly related to your question.  This paper is the closest thing I've found. It's concerned with the Euclidean plane, but the very beginning treats an arbitrary metric space.  
Let $A$ be a nonempty subset of a metric space $(X,d)$ and let $x\in X.$  Define the distance from $x$ to $A$ by $$D(x,A) = \inf\{d(x,a)|a\in A\}.$$  Let $A$ and $B$ be nonempty subsets of a metric space $X.$  Define the equisdant set determined by $A$ and $B$ as $$\{A=B\}=\{x\in X| d(x,A)=d(x,B) \}$$ 
In your question, $A$ and $B$ are singleton sets. 
In the paper it is shown that

Theorem $\mathbf{1}$ The metric space $X$ is connected if and only if equidistant sets in $X$ are never void.

In $\mathbb{Q}$, $\{A=B\}\neq\emptyset$ when $A$ and $B$ are singleton sets, but if we take $$A=\{x\in\mathbb{q}|x>\sqrt{2}\}\\
B=\{x\in\mathbb{q}|x<\sqrt{2}\}$$ then clearly, $$\{A=B\}=\emptyset.$$ 
A: Your additional conditions, mentioned in the edit, together with the condition from the comments that $X$ is actually a field, are almost sufficient for the existence of a point with equal distance. The only other thing we have to add is that the field is not of characteristic $2$ (that is, in that field $1+1\ne 0$).
To see this, first note that $d_X(x,0) = d_X(0,x) = d_X(-x,0)$ (the first by symmetry of the metric, the second by your second restriction, which BTW just says the metric is translation invariant).
Now define $c=\frac{a+b}{2}$; since $2\ne 0$, this element exists by the field axioms. Then we have
\begin{align}
d_X(a,c) &= d_X(a-c,0) && \text{(translation invariance)}\\
&= d_X(\frac{a-b}{2},0) && \text{(inserting $c$ and simplifying)}\\
&= d_X(\frac{b-a}{2},0) && \text{(the relation above)}\\
&= d_X(b-c,0) && \text{(using the definition of $c$ again)}\\
&= d_X(b,c) && \text{(translation invariance)}
\end{align}
