proofs:
We want to prove [tex]S \subset S \cup T[/tex]
Let [tex]s\in S[/tex]. By definition [tex]S\cup T[/tex] is the set that contains the elements of [tex]T[/tex] and the elements of [tex]S[/tex]. Then [tex]s[/tex] must be in [tex]S\cup T[/tex]. As [tex]s[/tex] was arbitrary, we conclude that [tex]S \subset S \cup T[/tex].
This proof is analogous to the previous one. In fact, this result is the same result as the previous one.
We want to prove [tex]S \cap T \subset S[/tex]
Let [tex]y\in S\cap T[/tex]. By definition of the intersection [tex]y[/tex] should be in [tex]S[/tex] and also in [tex]T[/tex]. Then, we already saw that [tex]y\in S[/tex]. As [tex]y[/tex] was arbitrary we can conclude that [tex]S \cap T \subset S[/tex].
This is the same result as the previous one. There is no need to prove it anymore, but if you wish, you can reply the exact same proof.
