If $S \subset T$ prove $\overline{S} \subset \overline{T}$ and $\text{int}(S) \subset \text{int}(T)$ Closure: $\bar{S} = \cap K$ where $K$ ranges over all closeds containing $S$.
Interior: $int(S) = \cup U$ where $U$ ranges over all opens contained in $S$. 
My attempt for first part: Let $x$ be an arbitrary point in $\bar{S}$. Then $x \in \cap K_S$. Note every element of $K_T$ is also in $K_S$ because every closed containing $T$ also contains $S$. So $\cap K_S$ = $\cap K_T \cap K_{S but not T}$. Now I just need to prove that this implies $\bar{S} \subset \bar{T}$ but I'm not sure how. Perhaps there is a set theory rule saying that: if $A$ and $B$ are sets of sets and $A \subset B$ then $\cap A \supset \cap B$?
My attempt for the second part: Clearly all open sets in $S$ are also contained in $T$. But not necessarily in the union of all open sets in $T$.... I just need to show that there exists an open set in $T$ that is not contained in $S$, and I'm not sure how. 
 A: To avoid confusion I am using the notation $\subseteq$ instead of $\subset$ preassuming that 'your' $S\subset T$ does not exclude that $S=T$.

Let $\mathcal K$ denote the collection of closed sets containing $S$.
As an intersection of closed sets the set $\overline{T}$ is closed, and this with $S\subseteq T\subseteq\overline{T}$. 
That means that $\overline{T}\in \mathcal K$ so that $\overline{S}=\cap \mathcal K\subseteq\overline{T}$.

Let $\mathcal V$ denote the collection of open sets contained in $T$.
As a union of open sets the set $\operatorname{int}(S)$ is open, and this with $\operatorname{int}(S)\subseteq S\subseteq T$. 
That means that $\operatorname{int}(S)\in \mathcal V$ so that $\operatorname{int}(S)\subseteq\cup \mathcal V=\operatorname{int}(T)$.
A: Another way to think about the closure of a set $X$ is $\overline{X} = X \cup X'$ where $X'$ is the set of limit points of $X$. We can now think about proving $$\overline{S} = S\cup S' \subseteq T \cup T' = \overline{T}$$ 

It is clear that $S \subset \overline{T}$. Now let $x \in S'$. For any open set $U$ containing $x$ then $S \cap [U\setminus\{x\}] \neq \emptyset$. By assumption that $S\subset T$ then $S \cap [U\setminus\{x\}]\subseteq T \cap [U\setminus\{x\}]$ so $ T \cap [U\setminus\{x\}]$ is nonempty. $U$ and $x$ were arbitrary, so $x$ is a limit point of $T$. We conclude $S' \subset T'$, and thus have the desired result $S\cup S' \subseteq T \cup T'$.


For your second proof, you do not want to show that there is an open set contained in $T$ but not in $S$. After all, you could have $T=S$. Instead, you could show that any $x\in \text{Int}(S)$ is also in $\text{Int}(T)$. 

Suppose FTSOC that there is some $x \in \text{Int}(S)$ but $x \notin \text{Int}(T)$. Then for every open set containing $x$, that set is not a subset of $T$. Notice that $\text{Int}(S)$ is an open set containing $x$, so $\text{Int}(S) \not\subset T$. This implies $S\not\subset T$, a contradiction. 

