Suppose that $*$ is an associative and commutative binary operation on a set $S$. Let
$$T = \{a \in S \, : \, a*a = a\}$$
Is one of these methods or both a correct approach in beginning to solve this problem?
To prove that $T$ is closed under $*$, I began with considering the object $a*b$, and then deduced that $a*b = (a*b)*(a*b)$, which implies that $a*b \in T$. But when I checked my solution, the instructor began with $(a*b)*(a*b)$ and deduced that $(a*b)*(a*b) = a*b$.
Here is my full solution. Let $a, b \in T$. By definition of $T$, we have
$$a = a * a \tag{1}$$ $$b = b * b \tag{2}$$
Consider the object $a*b$. Then because $*$ is commutative and associative and that $(1)$ and $(2)$ are true, we have
$$\begin{align} a*b &= (a*a) * (b*b) \\ &= a * [a * (b*b)] \\ &= a * [(a*b)*b] \\ &= a * [(b*a)*b] \\ &= a * [b * (a * b)] \\ &= (a*b) * (a*b) \end{align}$$
Therefore, $a*b = (a*b) * (a*b)$ and $a*b \in T$, as desired.
Both seem to work; equality is symmetric, so $(a*b)*(a*b)=a*b$ is equivalent to $a*b=(a*b)*(a*b)$.
However, it seems that perhaps your instructor's solution is the more "intuitive" direction, at least to me. Perhaps you could post your full solution?