showing that the pair of Finite Automata are equivalent
Your approach works. You are effectively showing that the symmetric difference $L(FA_1) \Delta L(FA_2)$ is empty, hence $L(FA_1) = L(FA_2)$.

A more direct, automaton-level approach would be to build the following product automaton $A = (Q, \Sigma, \delta, q_0, F)$:


$Q = Q_1 \times Q_2$
$\delta((q_1, q_2), a) = (\delta(q_1, a), \delta(q_2, a)) \ \forall (q_1, q_2) \in Q, a \in \Sigma$
$q_0 = (q_{0,1}, q_{0,2})$
$F = \{(q_1, q_2) \in Q \mid q_1 \in F_1 \Leftrightarrow q_2 \in F_2\}$


Intuitively, you keep in mind the current state in both $FA_1$ and $FA_2$. Initially, the current states are the respective initial states $q_{0,1}$ and $q_{0,2}$. Then, for each input symbol $a$, you move to the respective successor in each automaton, and note the successor state (the combination of the two successor states as a pair). Of course you only draw the state for each combination only once. You can proceed either breadth-first (considering all input symbols for one product state before moving on to a newly introduced state) or depth-first (always continue with the newly introduced state and backtracking if all product successor already existed).

Note that this way you might not have to construct the full product set $Q_1 \times Q_2$, as you automatically prune unreachable states (you don't ever introduce them). However, the product automaton in your example will have no more than 6 states anyway, so it is rather small.