Does compactness depend on the metric? If so, what are some examples of sets that are compact with respect to one metric but not compact with respect to another metric?
 A: Take $[0,1] $ with the usual metric, in which it is compact. And then consider it with the discrete metric, where $d (x,y)=1$ if $x\ne y$. In the latter, only finite sets are compact.
A: Compactness is a topological property, so if you have two metrics that induce the same topology, then either both metric spaces are compact, or else neither is compact.
However, if you have two metrics that are allowed to be topologically inequivalent, then surely one can be compact and the other one non-compact. For example the spaces $\left] 0,1 \right[$ and $[0,1]$ with their usual metrics are compact and non-compact, respectively, but the underlying sets of the spaces have cardinality $2^{\aleph_0}$ in both cases, so you could identify the underlying sets by some bijection of your choice, and say this is an example of the kind you ask for in your question.
More generally, let $A$ be any set.
If $|A|$ is finite, then any topology on $A$ makes $A$ compact. In particular any metric on $A$ makes a compact metric space. (Actually, any metric on a finite set induces the discrete topology, but even this topology is still "small" enough to be compact in this case.)
If $|A|$ is transfinite, you can choose the discrete metric $d(x,y)=1$ for all $x\ne y$. That gives one metric which is not compact. There will be other topologies on this set $A$ which give compact spaces (for example the trivial topology $\{ \varnothing, A \}$, or the cofinite topology) but they may not come from a metric (these topologies may be non-metrizable). I am not sure for which transfinite cardinalities $|A|$ we can choose a metric which makes the space compact, but as we saw, $|A| = 2^{\aleph_0}$ is one such cardinality.

As mentioned by user N. S. in a comment to an earlier answer, the case $|A|=\aleph_0$ is easy enough. One can pick a countable, closed and bounded subset of $\mathbb{R}$, such as $\{\frac1n \mid n\in\mathbb{N}\} \cup \{ 0 \}$. With the metric inherited from $\mathbb{R}$, this is a compact metric space of cardinality $\aleph_0$.
A: Take the set $[0, 1] \subset \mathbb{R}$ under the usual metric.  This is compact.  Now consider $[0, 1]$ instead under the discrete metric: $d(x, y) = 1$ if $x \neq y$ and $0$ otherwise.
This is not compact.  Can you see the infinite cover of open sets that does not admit a finite subcover?
This trick works with any compact infinite set, and further any infinite set with this so-called discrete metric cannot be compact.  On the other hand, any finite set with any metric is compact.
A: If you ask about arbitrary metrics, the other answers cover that well.
But if you ask about two metrics which define the same topology, then the answer is no, as it is made clear by the open cover definition of compactness.
P.S. Boundedness depends on the metric, and this shows that in general compactness cannot always be equivalent to bounded+closed.
