Then you should have said that! The set {1, 2} with the "indiscrete topology" (the only open sets are the empty set and entire set) is connected and contains exactly two points. Of course this topology is not metric.
In a metric space, given two points, p and q, there exist a continuous function mapping [0, 1] to the space such that f(0)= p and f(1)= q. Since the interval [0, 1] is uncountable, the image of the function is uncountable and so is any set containing it.
Let d denote the metric on M, and let the two given points in M be p and q. The function f given by f(x) = d(x,p) is continuous from M to the real numbers. Clearly f(p)=0 and f(q)=d(p,q)>0. The aim is to show that the range of this function contains the whole interval [0,d(p,q)].
In fact, if there exists a number r with 0<r<d(p,q) such that f(x) is never equal to r, then the sets U={x : f(x)<r} and V={x : f(x)>r} are nonempty, open, disjoint, and their union is the whole of M. That contradicts the connectedness of M. Therefore the range of f contains the (uncountable) interval [0,d(p,q)], from which it follows that M is uncountable.