Prove: If [S ⊆ T], then (S ∩ T) ⊆ S ⊆ (S ∪ T) I'm new to this entire proof thing, and I am so confused.
This question basically says
[S ⊆ T] = [∀xES,(xET)]
Prove that for two arbitrary sets S, and T,
(S ∩ T) ⊆ S ⊆ (S ∪ T)
If you can try to dumb it down so I can understand please!
 A: One way for the proof would be:
Let $x\in S\cap T$, so according to the definition we have $$x\in S\wedge x\in T$$ so $x\in S$ and therefore $S\cap T\subseteq S$. Also we know that $$S\cup T=\{x\mid x\in S\vee~ x\in T\}$$, so it is obvious that we have $S\subseteq (S\cap T)$. However $S\subseteq T$ makes us to have: $$S\cap T=S,~~S\cup T=T$$
A: Recall that for any two sets $A$ and $B$ the intersection ($\cap$) of them is defined to be
$$
A\cap B:=\{x\mid x\in A\;\text{and}\;x\in B\}
$$
and the union ($\cup$) of them is defined as
$$
A\cup B:=\{x\mid x\in A\;\text{or}\;x\in B\}.
$$
In general if you want to prove that $A\subseteq B$ you might do it by "element-chasing". That is: Let $x\in A$ be an arbitrary element of $A$. Then show that $x\in B$ as well. 
As for your question, you want to prove that for any two sets $S$ and $T$ we have $$S\cap T\subseteq S\subseteq S\cup T.$$ So we have two statements to show: (1) $S\cap T\subseteq S$ and (2) $S\subseteq S\cup T$. 
For (1) we let $x\in S\cap T$ be an arbitrary element of $S\cap T$. By definition this means that $x\in S$ and $x\in S$ and hence we conclude that $x\in S$. Thus we have shown that $S\cap T\subseteq S$.
For (2) we argue similarly.
A: Suppose that $x\in S\cap T$. Then, both $x\in S$ and $x\in T$ by the definition of intersection. In particular, $x\in S$. Therefore, $x\in S\cap T$ implies $x\in S$, so that $S\cap T\subseteq S$.
Now, suppose that $x\in S$. It is true that either $x\in S$ or $x\in T$ (or both), so that, by the definition of union, $x\in S\cup T$. Therefore, $x\in S$ implies that $x\in S\cup T$, so that $S\subseteq S\cup T$.
