In a general metric space, any finite subset (with more than one element) is always disconnected. To see this, suppose $X\subset M$ is a finite subset of some metric space with metric $d$, with more than one element. Let $\alpha=\min\{d(a, b): a, b\in X, a\not=b\}$; since $X$ has more than one element, this makes sense, and we have $\alpha>0$ (why?). Now, think about the balls around points in $X$ of radius $\alpha\over 2$ . . .
More generally, what's going on is that metric spaces are Hausdorff, and any finite Hausdorff space is disconnected (or has only one point). Note that there are topological spaces (necessarily non-metrizable) in which finite subspaces with more than one element can be connected; for example, consider the topology $\{\emptyset, \{1\}, \{0, 1\}\}$ on $\{0, 1\}$.