Solution: Let $A$ is dense in $X$, that is, $\bar{A} = X$. We need to show that $A$ is dense in $B$ and $B$ is dense in $X$. Note that the closure of $A$ in $B$ is $B \cap \bar{A} $. So, we have \[ B \cap \bar{A} = B \cap X = B. \] Thus, $A$ is dense in $B$. Also, observe that \[ A \subseteq B \subseteq X \implies \bar{A} \subseteq \bar{B} \subseteq \bar{X} \implies X \subseteq \bar{B} \subseteq X \implies \bar{B} = X. \] Thus, $B$ is dense in $X$.

Conversely, let $A$ is dense in $B$ and $B$ is dense in $X$. Then we have \[ \bar{A} \cap B = B \text{ and } \bar{B} = X. \] Since \[ \bar{A} \cap B = B \implies B \subseteq \bar{A} \implies \bar{B} \subseteq \bar{A} \implies X \subseteq \bar{A} \implies \bar{A} = X. \]