I wish to add an additional and slightly tangential answer about the term "T-finite".
Tarski proved that a set $S$ is finite if and only if every non-empty collection of subsets of $S$ has a maximal element. This equivalence holds without the axiom of choice, and so it made somewhat sense to slightly weaken it during the years when various forms of finiteness were investigated (without the axiom of choice, of course).
In Jech's book The Axiom of Choice he defines (Ch. 4, Ex. 9, p. 52) T-finite in a slightly different manner:
Call a set $S$ $T$-finite if every non-empty monotone [read: $\subseteq$-chain] $X\subseteq\mathscr P(S)$ has a $\subseteq$-maximal element.
It is not hard to show that every finite set is $T$-finite, indeed Tarski's equivalent implies that immediately. It is also not hard to show that every $T$-finite is Dedekind-finite.
Neither of the implications is reversible in ZF, as the following will show:
If $A$ is amorphous (infinite and every subset of $A$ is finite or co-finite) then $A$ is $T$-finite. It is consistent that amorphous sets exist. Therefore it is consistent that there are $T$-finite sets which are infinite.
If $A$ is a $T$-finite set then $A$ can be linearly ordered if and only if $A$ is finite.
It is consistent that there is an infinite Dedekind-finite set of real numbers. The previous fact shows that such set cannot be $T$-finite because it can be linearly ordered.
For those interested, a nice exercise in understanding the definition is to show that we can replace "maximal" by "minimal" in all the definitions.