If $A$ is dense in $S$ and $S$ is dense in $T$ , then $A$ is dense in $T$ In a metric space $M$, if the subsets satisfy $A \subseteq S \subseteq \bar A$ where $\bar A$ refers to the closure of $A$, then $A$ is said to be dense in $S$. If If A is dense in S and S is dense in T , show that then A is dense in T.
Attempt: Given that $A$ is dense in $S \implies A \subseteq S \subseteq \bar A$ 
and $S$ is dense in $T \implies S \subseteq T \subseteq \bar S$.
We need to prove that $S \implies A \subseteq S \subseteq T \subseteq  \bar A$.
Let us suppose that $A \subseteq S \subseteq  \bar A \subseteq T \subseteq \bar S 
\quad\ldots(1)$.
Then it can be inferred that $A $ is dense in $S$, $S$ is dense in $\bar A$, $S$ is dense in $T$.
Hence, $\bar A$ might contain some accumulations points of $S$ . $T$ might also contain some accumulation points of $S$.
How do I proceed to bring about a contradiction to assumption $(1)$?
Thank you for your help.

Remember that $S\subset \bar{A}$ implies $\bar{S}\subset\bar{A}$
