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.