In Naive Set theory by Halmos, he states $a \in A$ is equivalent to $ \{ a \} \subset A $. I am guessing that here he means logically equivalent. Intuitively, I see how they can be interchanged, but I am not sure how to prove it.
I attempted to set up a truth table. I tried breaking down $B \subset A$ into the sentence $ \forall x(x \in B \implies x \in A) $.
I know that truth tables in predicate logic would be infinitely long, but I thought that in this special case, since we only have $\{ a \}$ I could employ a truth table with $B = \{ a \}$ and $x = a$.
$$\begin{array}{c|c|c|} a \in A & a \in B & \implies & a \in A\\ \hline \bf{T} & T & \bf{T} & T \\ \hline \bf{T} & F & \bf{T} & T\\ \hline \bf{F} & T & \bf{F} & F\\ \hline F & F & T & F\\ \hline \end{array}$$
This truth table shows that these are not logically equivalent. What am I doing wrong?
You went wrong in assuming that this is a logical equivalence. It is not; set-theoretically they are equivalent, but logically they are not, as you yourself demonstrated.
Specifically, $a \in \{ a \}$ has to be true set-theoretically (and hence the last two rows of the truth-table do not apply), but it is not a logical truth.
Put a different way: once you assume some basic axioms regarding set-theory (in particular ones that define the notion of subset ... or what something like a singleton set means) you can derive $\{ a \} \subset A$ from $a \in A$, and vice versa. But without those axioms, you cannot. So that shows that these are not logically equivalent, but 'merely' equivalent in the context of set-theory.