Simple Set Theory Proof

Hi - I'm taking a linear algebra course and the first few chapters are on set theory. I'm having trouble with a very simple proof and I'd like some help on it.

Let $S$ and $T$ be groups. Given $T \subseteq S$ prove:

a) $S \cap T = T$

b) $S \cup T = S$

I know these are very simple proofs but please keep in mind (and I know it's ridiculous) I was never taught proofs in high school, so the concept of proving something quite obvious is new to me.

Here's my shot at it:

a) Let $x \in T$

  - And since $T \subseteq S$, then $x \in S$

  - Which means $x \in S \cap T$

  - And of course, $S \cap T \subseteq S$

  - And since $S \cap \{ x | \{x\} \subseteq S\} = \{x | \{x\} \subseteq S\}$

  - It is implied that $S \cap T = T$

b) Let $x \in T \ and \ y \in S$

  - Since $S \cap T = \{x,y \ | \ x \in T \ , \ y \in S \}$ and it is given that $T \subseteq S$

  - And we already know that $x \in T \Rightarrow x \in S$

  - Then the previous statement can be rewritten as $S \cup T = \{x,y \ | \ x \in S , y \in S \}$

  - Which obviously implies that in fact $S \cup T = \{ x \ | \ x \in S\}$

  - So $S \cup T = S$

Can anyone tell me if these proofs are correct and if not, explain them in similar terms so I can improve my strategies?