As mentioned in several replies, you require that the compact space be metrisable for a positive answer to your question.  Then the claim is, in fact, true.  The natural setting for this question is the Banach-Dieudonn\'e theorem.  If $E$ is a separable Banach space, then, as has been mentioned in previous answers, the  unit ball of the dual space is metrisable under the weak $*$-topology.  The natural topology on the dual is then the finest locally convex topology on the whole space which agrees with the latter on this ball---it is even the finest such topology (i.e., not necessarily locally convex or even linear) and can also be characterised   as the topology of uniform convergence on compact subsets of $E$.  This topology has several nice properties---it is complete and has the same bounded subsets as the norm topology---but it is not metrisable.  However, its restrictions to the bounded sets of $E'$ are metrisable.
This is the essential content of the above-mentioned result.  The consequence which is relevant to the question posed is the fact that  it has the same convergent sequences as the weak $*$-topology on $E'$ defined by a dense countable subset of $E$ and this {\it  is} metrisable.  In order to answer the question posed, it suffices to specialise to the case where $E$ is $C(K)$ with $K$ compact and metrisable.

I should perhaps mention that the theorem of Banach-Dieudonn\'e holds for any (i.e., not necessarily separable) Banach (or even, in suitable form, Fr\'echet) space and this provides information for the case where $K$  is not metrisable.