1. **Problem statement:** Given two non-empty subsets $S$ and $T$ of a topological space $(X, \tau)$ such that $S \subseteq T$, show that if $S$ is dense in $X$, then $T$ is also dense in $X$. 2. **Recall the definition of density:** A subset $A$ of $X$ is dense in $X$ if the closure of $A$ equals $X$, i.e., $$\overline{A} = X.$$ This means every point in $X$ is either in $A$ or is a limit point of $A$. 3. **Given:** $S$ is dense in $X$, so $$\overline{S} = X.$$ Also, $S \subseteq T$. 4. **Use the property of closures:** For any subsets $A$ and $B$ of $X$, if $A \subseteq B$, then $$\overline{A} \subseteq \overline{B}.$$ Applying this to $S \subseteq T$, we get $$\overline{S} \subseteq \overline{T}.$$ 5. **Substitute the known closure:** Since $\overline{S} = X$, it follows that $$X \subseteq \overline{T}.$$ But $\overline{T}$ is always a subset of $X$, so $$\overline{T} = X.$$ 6. **Conclusion:** Since the closure of $T$ is $X$, $T$ is dense in $X$. **Final answer:** If $S$ is dense in $X$ and $S \subseteq T$, then $T$ is dense in $X$.