If $A$ and $B$ are both finite, it suffices to show:

(i) $A \subseteq B$ and (ii) $\# A = \geq \#B$.

There are variations on this when the sets have more structure e.g.:

If $A$ and $B$ are finite-dimensional vector spaces over a field $K$, it suffices to show that one is contained in the other and that they have the same dimension.

For "naked" infinite sets, I am tempted to say that I know of no way to show equality other than directly from the definition: every element of $A$ is an element of $B$ and conversely. So I'm interested to hear what others have to say. [Edit: I feel that Qiaochu's answer successfully meets this challenge: it is another way to show equality, and it is both obvious and useful.]

If $A$ and $B$ are both finite, it suffices to show:

(i) $A \subseteq B$ and (ii) $\# A = \#B$.

There are variations on this when the sets have more structure e.g.:

If $A$ and $B$ are finite-dimensional vector spaces over a field $K$, it suffices to show that one is contained in the other and that they have the same dimension.

For "naked" infinite sets, I am tempted to say that I know of no way to show equality other than directly from the definition: every element of $A$ is an element of $B$ and conversely. So I'm interested to hear what others have to say.