Help with Subset Problem So I'm "supposed to show that":
$(A \cup B)  \subseteq (A \cup B \cup C)$.
For $(A \cup B)$, $x \in A \cup B$ and therefore $x \in A \lor x \in B$. For
$(A \cup B \cup C )$ , $x \in A \cup B \cup C$ and therefore $x \in A \lor x \in B \lor x \in C$. I'm not quite sure how to explain my results.  The best I can come up with is because $x$ is an element of set $A$ or set $B$ or set $C$,  $(A \cup B)$ is a subset of $(A \cup B \cup C)$. How can I explain this better?
Follow up question:
Thanks for the explanations and I'd like to get help on another one.
$(A \cap B \cap C) \subseteq (A \cap B)$.
So for $(A \cap B)$, $x \in A \land x \in B$. And $x \in A$ and $ x \in B$ must be true. Can I assume that  $(A \cap B \cap C)$ must be true as well if it is a subset of $(A \cap B)$ and that $x \in C$ must be true?
 A: HINT: Note that $Y\subset X$ if and only if $Y\cap X=Y $
A: To show that $P\subset Q$, you suppose that $x$ is an element of $P$, and show that it must also be an element of $Q$.
Here you suppose $x \in A\cup B$, so, as you said, you know either $x\in A$ or $x\in B$.  You don't know which, so you proceed by cases.


*

*Suppose $x\in A$. Then $x \in A\cup B\cup C$ because... ?

*Suppose $x\in B$. Then $x \in A\cup B\cup C$ because... ?

A: You have to remember that if you have two statements $p$ and $q$, then the statement $p \lor q$ is true if both $p$ and $q$ are true, or if one is true and one is false.  The only time $p \lor q$ is a false statement is when both $p$ and $q$ are false statements.
So, to prove $A \cup B \subseteq A \cup B \cup C$, you first let $x \in A \cup B$, as you did.  This means either $x \in A$ is true, or $x \in B$ is true (or both!).
But since at least one is true, that means the statement $(x \in A) \lor (x \in B)$ is a true statement (as we just said above).  But if this is a true statement, then the statement $[(x \in A) \lor (x \in B)] \lor x \in C$ is always true, because we just established that the first piece $[(x \in A) \lor (x \in B)]$ is true.  
But the parentheses don't matter, I just put them in there for you to see what I'm saying.
So, we end up with $x \in A \lor x \in B \lor x \in C$ is a true statement, which is what it means for $x \in A \cup B \cup C$ to be true, and that's it.
A: The logic of a proof must proceed from the premises to the conclusion. You have the right start:

For $(A \cup B)$, $x \in A \cup B$ and therefore $x \in A \lor x \in B$.

However, using

For $(A \cup B \cup C )$ , $x \in A \cup B \cup C$ and therefore $x \in A \lor x \in B \lor x \in C$.

to continue is not correct. You have the right ideas here. It's crucial to get your statements in the correct order to have a valid proof.  One way to figure out the correct order is to work from what you are trying to prove:

In order to show that $x \in (A \cup B \cup C )$, I need to first show that $x \in A \lor x \in B \lor x \in C$.

The rules of logic tell us that that $x \in A \lor x \in B$ implies $x \in A \lor x \in B \lor x \in C$.
Now we write this thought process as the final proof:

Let $x \in A \cup B$ and therefore $x \in A \lor x \in B$. Thus $x \in A \lor x \in B \lor x \in C$. Therefore, $x \in A \cup B \cup C$.

For your second proof, follow this same procedure. Start with "Let $x \in (A \cap B \cap C)$" and follow the logic of the definitions. Also work from the other end by observing that to show $x \in (A \cup B)$ you need to show that $x \in A \land x \in B$.
