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This is my first question on this site, so i'm going to start with presumably an easy one.

To show a subset, is it valid to use a contrapositve proof. For example I want to show that $A \subset B$, can I assume $ x \notin B$ which will eventually imply that $ x \notin A$.

I feel that this works, however I have a subconscious voice telling me that I have overlooked something.

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    To show containment, you show that: if $x\in A$ then $x\in B$. The contrapositive of this is (as you said): if $x\notin B$ then $x\notin A$. If the contrapositive is easier to prove (or even if it's not), you can certainly do this (the contrapositive of a statement is equivalent to the original statement). Long story short: yes, you can do this. – 2017-02-12
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    Ok, so this is officially the best thing ever. You will be hearing a lot from me. I was trained as a physicist, but decided to get a masters in mathematics to bone up on my skills before pursuing a PHD in theoretical physics. However , in my young arrogance as a physicist I brushed set theory to the side and its really coming back to haunt me. – 2017-02-12
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    i) $x \in A \implies x\implies B $ if and only if 2) $x\not \implies non B \implies x \not \implies nanny A $. So yes. Either prove $A\subset B $. But 2) will probably (not certainly but probably) be much more difficult to prove. But you can certainly attempt 2) if you like. – 2017-02-12

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That's right! To say $A \subset B$ is to say that $x\in A$ implies $x \in B$.
Contrapositively, that is $x \notin B$ implies $x \notin A$.
Here we are saying that the only way for $A$ not to be a subset of $B$ is if $A$ contains something that $B$ doesn't: that is, $x \in A$ but $x \notin B$.

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Yep, that works! The statement $A\subset B$ just means that if $x\in A$, then $x\in B$. Like any implication, it is equivalent to its contrapositive, which is that if $x\not\in B$ then $x\not\in A$.