1. **Problem Statement:** Given two non-empty subsets $S$ and $T$ of a topological space $(X, \tau)$ such that $S \subseteq T$, and a point $p$ which is a limit point of $S$, we need to verify that $p$ is also a limit point of $T$ and deduce that the derived set $\overline{S}$ (set of limit points of $S$) is a subset of $\overline{T}$ (set of limit points of $T$). 2. **Recall the definition of a limit point:** A point $p \in X$ is a limit point of a set $A \subseteq X$ if every open neighborhood $U$ of $p$ contains at least one point of $A$ different from $p$ itself. 3. **Given:** $S \subseteq T$ and $p$ is a limit point of $S$. 4. **To prove:** $p$ is a limit point of $T$. 5. **Proof:** Since $p$ is a limit point of $S$, for every open neighborhood $U$ of $p$, we have $$U \cap (S \setminus \{p\}) \neq \emptyset.$$ Because $S \subseteq T$, it follows that $$U \cap (T \setminus \{p\}) \supseteq U \cap (S \setminus \{p\}) \neq \emptyset.$$ Therefore, every open neighborhood $U$ of $p$ contains a point of $T$ different from $p$, so $p$ is a limit point of $T$. 6. **Deduction:** Since every limit point of $S$ is also a limit point of $T$, the derived set of $S$ is a subset of the derived set of $T$: $$\overline{S} \subseteq \overline{T}.$$