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 for bounded sequences. 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 star 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 bounded convergent sequences as the weak star 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.

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 for bounded sequences. The natural setting for this question is the Banach-Dieudonné 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 star 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 bounded convergent sequences as the weak star 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-Dieudonné holds for any (i.e., not necessarily separable) Banach (or even, in suitable form, Fréchet) space and this provides information for the case where $K$ is not metrisable.

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.

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 for bounded sequences. 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 star 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 bounded convergent sequences as the weak star 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.