In proving A = B, A, B are sets, do you always have to show $\subseteq$ and $\supseteq$? I am trying to show the DeMorgan's Law 

$X \backslash \bigcup_{\alpha \in I} A_\alpha = \bigcap_{\alpha \in I}
(X \backslash A_\alpha)$

It seems I could directly approach this as follows:
$X \backslash \bigcup_{\alpha \in I} A_\alpha = X \bigcap (\bigcup_{\alpha \in I} A_\alpha)^c  = X \bigcap (\bigcap_{\alpha \in I} A_\alpha^c) = \bigcap_{\alpha \in I} X \backslash A_\alpha$
The last line follows from distributivity
But I thought in set theory proofs of the type Prove $A = B$, you have to show that $ A \subseteq B$ and $B \subseteq A$. 
But in this case it seems I have directly proved that the two are equivalent without resorting to $A \subseteq B$ and $B \subseteq A$...
Can someone enlighten me as to whether my approach is correct? I am not very well versed in set theory proofs. 
 A: Your proof is just fine. :-)
The rule "to show $A = B$, verify that $A \subseteq B$ and that $B \subseteq A$" is quite nice, since it often gives you a fruitful starting point for your proving efforts.
However, it's not actually necessary to verify an equality using this rule. Any other sound method of proof can be used as well. In your case, you implicitly used the lemma "if $A = B$ and $B = C$ then $A = C$".
Let me add that calculational proofs proceeding by stringing equalities might be considered more elegant than element-wise proofs. However, in many cases, element-wise proofs are easier to accomplish.
A: There is no rule that one must show inclusion both ways in a set theoretical proof for the equality of two sets, it is just very often an easy way to do it. If you can do it without, by all means. 
I do question your step $X \bigcap (\bigcup_{\alpha \in I} A_\alpha)^c  = X \bigcap (\bigcap_{\alpha \in I} A_\alpha^c)$ since you are using the fact that $(\bigcup_{\alpha \in I} A_\alpha)^c  = \bigcap_{\alpha \in I} A_\alpha^c$, which is just another formulation of De Morgan's rule itself. If you are trying to prove De Morgan's rule in general, you can't use that fact without having a circular argument. However, if you are just proving another formulation of De Morgan's rule and you are allowed use this formulation, then everything is fine.
A: A direct proof : For all $y$ we have $$y\in X\backslash \cup_{a\in I}A_a \iff (y\in X \land y\not \in \cup_{a\in A}A_a)\iff$$  $$\iff   (y\in X \land (\forall a\in A\;(y\not \in A_a)))  \iff (\forall a\in A\; (y\in X \land y\not \in A_a)) \iff $$ $$\iff (\forall a\in A (y\in X\backslash A_a))\iff y\in \cap_{a\in A}(X\backslash A_a).$$
In some problems it may be easier to show $A\subset B$ and $B\subset A$ to prove A=B. In many cases,one half is easier than the other, or different methods may  be more suitable for the two halves.
