I already proved that if a set A is compact in a metric space M = (X,d) then A is closed and bounded, however, can we say that if I have a set A that's closed and bounded it's also compact in M?
(1). Take any metric space $(X,e).$ Let $d(p,q)=\min (1,e(p,q)).$ Then $(X,d)$ is a metric space. And the metrics $e,d$ generate the same topology on $X.$ Compactness is a property of the topology. So if $(X,e)$ is not a compact metric space then $(X,d)$ is not. But $X$ is a closed bounded subset in the space $(X,d).$
(2). Take any metric space $(Y,d)$ that has a bounded non-closed subset $X.$ The subspace $(X,d)$ is not compact, and $X$ is a closed bounded subset in the space $(X,d).$... (although it is not closed in the space $(Y,d)$.)
(3). Your title suggests a different interpretation of the Q. A topological space $A$ is metrizable iff its topology can be generated by a metric, and any metric that generates that topology is called a metric for $A.$ If $A$ is metrizable and compact then any metric for $A$ is bounded.
And if $A$ is metrizable and non-compact then there exists an unbounded metric for $A.$ There are many ways to prove this.
No. Take $A = M$, which is always closed, so you're asking if a bounded metric space is always compact. This is false and there are many counterexamples; for example, consider an infinite set with the discrete metric, where two non-identical points are at distance $1$ from each other.