Yes, "compactness" does depend strongly on the metric. A simple example is the metric d(x,y)= 1 if x\ne y, 0 otherwise. That is called the "discrete metric" because using that metric on any set gives the "discrete topology" in which all sets are open. To see that, recognize that the "neighborhood of point p with radius 1/2" is just the singleton set {p} itself- since the distance from p to any other point is 1, the only point, x, for which it is true that d(p, x)< 1/2 is p itself. Every point, in every set, is an interior point because that singleton set is a subset neighborhood contained in the set.
If X is an infinite set, with the discrete metric, then the compact sets are exactly the finite sets. That is because we could always use the individual singleton sets, {a} where a is any member of A, as our open cover. Since every point of A is in only one of those, we cannot remove any of them, much less reduce to a finite cover.
So, for example, take X to be the set of all real numbers.
With the "usual metric", d(x,y)= |x- y|, the interval [0, 1] is compact because it is closed and bounded. With the discrete metric, it is not compact because it is not finite.
Another example: X is the set of real numbers formed by the sequence {0, 1, 1/2, 1/3, ..., 1/n, ...}, again with the usual metric on the real numbers. Let \{U_n\} be any open cover. Since 0 is in A, there exist some U_0 which contains 0. Since U_0 is open, 0 is an interior point- there exist \delta such that \{x | |x|< \delta\} is a subset of U_0. But the sequence 1, 1/2, 1/3, ... converges to 0. There exist some N such that if n> N, |1/n|< \delta and so all 1/n, for n> N, is in U_o. Pick a single U_n that contains 1/n for n< N. That is a finite collection and it, together with U_0, makes a finite subcollection that covers A. A is compact.
But that same set of real numbers, with the discrete metric, is infinite and so is not compact.
