 Solution

True. If $\mathrm{TC}(\{x\})$ is countable, then each of its members is a subset of this transitive set and is therefore countable.

Conversely, if $x$ is reasonable, then $x$ itself and every set occurring at every finite membership depth below it are countable. Starting with the countable set $\{x\}$, each next level is a countable union of countable sets, hence countable. Their countable union is $\mathrm{TC}(\{x\})$, so the transitive closure is countable.