TITLE: Singleton subset of a metric space
QUESTION [8 upvotes]: I am currently working through chapter two of Principles of Mathematical Analysis (ed. 3) by Walter Rudin. My question comes from pages 30-31.
I know that a metric space must satisfy the definition:

Definition: A metric space is an ordered pair, $(X,d)$, where $X$ is a set (whose elements are called points), and $d$ is a metric. The metric must satisfy the following, given $x,y,z \in X$:
(a): $d(x,y) \ge 0$
(b): $d(x,y) = 0$ iff $x=y$
(c): $d(x,y) =d(y,x)$
(d): $d(x,z) \le d(x,y) + d(y,z)$

Rudin then provides the example that euclidean spaces, $\mathbb{R}^k$ are also metric spaces.
Following, he then specifies that any subset of a metric space is also a metric space.
After proving this statement, I then decided to come up with examples, and as an exercise, try to challenge myself to come up with counter-examples. One attempt that I made was: $\{ \boldsymbol{x} \} \subset \mathbb{R}^k$ where $\boldsymbol{x}$ is a $k$-tuple. This provides the conjecture:

Conjecture: Given any $\boldsymbol{x} \in \mathbb{R}^k$, the set $\{ \boldsymbol{x} \} \subset \mathbb{R}^k$ is a metric space.

Proof: $d(x,x) = |x - x| = |0| = 0$. Therefore, the only possible metric is $0$ and hence all the definition's conditions are satisfied.


I am uneasy about the proof I have constructed because in the definition, two (and once three) elements are drawn from the metric space. So in order for the above to hold, I essentially need to give this element several times. 
So the primary substance of my question is: in a metric space, can we let an element "do two jobs at once?"

REPLY [4 votes]: This would be another examples in which you can see @Brian's remarks are ruling again. Let $X\ne\emptyset$ and $$
d(a,b) = \left\{
        \begin{array}{ll}
            0 & \quad a=b \\
            1 & \quad a\ne b
        \end{array}
    \right.~~~~~~\text{or}~~~~~~d(a,b) = \left\{
        \begin{array}{ll}
            2 & \quad a\neq b \\
            0 & \quad a= b
        \end{array}
    \right.
$$