Showing that a finite or countable set in $\mathbb{R}^k$ is not connected I have been using this result and I've looked in several books that all state this result but don't give a proof:

Any finite or countable set in $\mathbb{R}^k$ is not connected.

Can anyone please explain? 
 A: I'll give a fourth answer that tries to integrate various components of the given arguments. Suppose a set $X$ is non-empty, at most countable and is connected. We aim to prove $X$ contains only a single point. Take some $x \in X$ and define $D = \{ d(x,y) : y \in X \}$.The function $f(y) = d(x,y)$ is continuous on $\mathbb{R}^k \to \mathbb{R}$ so the image is connected in $\mathbb{R}$. Connected sets on the line are intervals, and the only at most countable interval is a point. So $D = \{0\}$ and $X$ is a single point. 
A: Here's a (perhaps dull) method that directly uses the definition.
Suppose the set $X \subset \mathbb{R}^k$ has at least two elements but is at most countable. Pick any two elements $x, y \in X$; for some $j \in \{1, \ldots, n\}$ the $j$th coordinates of $x, y$ are different. By relabeling if necessary, for convenience we can suppose $x_j < y_j$.
Now, let $\pi: \mathbb{R}^k \to \mathbb{R}$ denote projection onto the $j$th coordinate (which is continuous). Since $X$ is countable, so is the subset $\pi(X)$; in particular the set difference $(x_j, y_j) - \pi(X)$ is nonempty, and so we can pick some $z$ in this set. So, $\pi^{-1}(\{z\}) \cap X = \emptyset$ and hence
$$U := \pi^{-1}((-\infty, z)) \cap X \qquad \text{and} \qquad V := \pi^{-1}((z, \infty)) \cap X$$
is a separation of $X$, as $U, V$ are both open and nonempty (by construction $x \in U$ and $y \in V$), $U \cup V = X$, and $U \cap V = \emptyset$.
A: suppose your set $X$ is connected: if you take any two disjoint open sets $U,V\subseteq \mathbb{R}^k$ such that
$$
X = (X \cap U) \cup (X \cap V)
$$
then one of these intersections would have to be empty. Now let
$$
U_r := \{ x | d(x,0)<r\} \qquad V_r := \{ x | d(x,0)>r\}
$$
since $D:=\{d(0,x)|x\in X\}\subseteq \mathbb{R}$ is countable or finite, it is not an interval. You can therefore find an $r>0,r\notin D$
such that there are $x_1,x_2 \in X$ with $d(0,x_1)<r<d(0,x_2)$ (here you need $X$ having more than one point) which is impossible.
A: Here's an another way to prove the result using a basic trick in these kinds of questions. (The basic idea is the same as in Blah's answer.) 
Suppose $X$ is at most countable and has at least two points $x$ and $y$. Since $X$ is at most countable, only countably many of the sets
$ X \cap S(x,r) $
can be non-empty. Since there are uncountably many radii $r$ such that $0 < r < d(x,y)$, there must exist a radius $r_0 \in ]0,d(x,y)$ such that $X \cap S(x,r_0) = \emptyset$. Now the sets
$$ A = X \cap B(x,r_0) \text{ and } B = X \setminus \overline{B}(x,r_0)$$
form a separation of $X$. (They are both open since $X \cap S(x,r_0) = \emptyset$.)
