I'm trying to prove that $(\Bbb Z/3\Bbb Z,+)$ is closed with respect to $+$. My first thought was to use the definition of the set $\Bbb Z/3\Bbb Z$ which is $\Bbb Z/3\Bbb Z=\{[0]_{3},[1]_{3},[2]_{3}\}$
A set $S$ is closed if $\forall a,b\in S (a\star b) \in S$ for some equipped binary operation $\star$.
The possible combinations for values of a and b are:
$a=3k, b=3k$
$a=3k,b=3k+1$
$a=3k,b=3k+2$
$a=3k+1,b=3k+1$
$a=3k+1,b=3k+2$
$a=3k+2,b=3k+2$
So then I would do the addition of all of these pairs and see if they come out in the form of either $3k,3k+1$, or $3k+2$ and if one of them doesn't, then the set is not closed. Is this the right approach?
That is right but you should note that any integer is of one (and only one) of the forms $3k, 3k+1$ or $3k+2$ by the division algorithm.