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.
 A: Remember that $S\subset \bar{A}$ implies $\bar{S}\subset\bar{A}$
A: (Note: Someone in comments below dislikes this answer because it assumes the space is metrizable.  But the question explicitly said it is a metric space.)
Let $t$ be a point in $T$ and let $\varepsilon>0$.  It suffices to show the open ball of radius $\varepsilon$ about $t$ contains at least one point of $A$.  Since $S$ is dense in $T$, that ball contains at least one point $s$ of $S$.  Now let $\eta=\varepsilon-d(t,s)>0$. Then the ball of radius $\eta$ about $s$ must contain some point $a$ in $A$.  Now use the triangle inequality to show that $d(t,a)<\varepsilon$.
A: You are given $A\subset S\subset\bar A$ and $S\subset T\subset\bar S$, and you must show $A\subset T\subset\bar A$.
From the hypotheses, you have $A\subset S$ and $S\subset T$, so $\boxed{A\subset T}$.
From the hypotheses, $T\subset\bar S$; you also have from the hypotheses that $S\subset\bar A$, so $\bar S\subset\overline{(\bar A)} = \bar A$. Thus $\boxed{T\subset\bar A}$.
The boxed conclusions above demonstrate that $\boxed{A\subset T\subset\bar A}$ as required. $\blacksquare$
