Choose $\displaystyle x_0\in \mathcal{M}$. If it is an isolated point then $\displaystyle \{x_0\}$ is open and closed, so $\displaystyle \mathcal{M}$ is not connected. Otherwise, let $\displaystyle D = \{d(x,x_0):x\in\mathcal{M},\;x\ne x_0\}$. Then D is a countable set of positive real numbers, with $\displaystyle \inf D = 0$. It may happen that D is not bounded above, in which case define $\displaystyle \sup D = \infty.$ In any case, we can find a real number $\displaystyle r\notin D$ with $\displaystyle 0<r<\sup D.$ Let $\displaystyle U = \{x\in\mathcal{M}:d(x,x_0)<r\}$, $\displaystyle V = \{x\in\mathcal{M}:d(x,x_0)>r\}$. Then $\displaystyle U,\;V$ are disjoint nonempty open subsets of $\displaystyle \mathcal{M}$ with $\displaystyle U\cup V =\mathcal{M}$. So $\displaystyle \mathcal{M}$ cannot be connected.
It is essential for that argument that $\displaystyle \mathcal{M}$ should be a metric space.
This example shows that a countable Hausdorff topological space can be connected.