Example of a set that is Dedekind-finite but not Tarski-finite? Can you give an example of a set that is Dedekind-finite, but not Tarski-finite?
 A: First let me set the definitions, so there will be no fuss about the terms I use later on.


*

*Finite means bijectible with a finite ordinal.

*Dedekind-finite means "every self-injection is a bijection" (or equivalently, every self-injection is surjective).

*Tarski-finite means that for every non-empty set of subsets has a $\subseteq$-maximal element.

*Whenever "infinite" is being used, possibly with one of the above modifiers, the meaning is that the set does not satisfy the relevant definition of being finite (e.g. Dedekind-infinite means not Dedekind-finite).


It is provable in $\sf ZF$ that every Tarski-finite set is finite. However it is consistent with the failure of the axiom of choice that there is an infinite set which is Dedekind-finite.
But because this is a consistency result, the word "example" is misleading. Since we cannot prove, nor disprove this, from the axioms of $\sf ZF$, giving a concrete example is impossible.
In the model defined by Paul Cohen, in which the real numbers cannot be well-ordered, he does that by adding an infinite Dedekind-finite set of real numbers. Since Tarski-finiteness is the same as finiteness, this is an "example" as you are asking for. But it cannot be "given explicitly" because this is a consistency result, rather than a provability result.

One can weaken the definition of Tarski-finite, and require that only $U$ which is linearly ordered by $\subseteq$ will have a maximal element; and under the axiom of choice this is again equivalent to being finite (and Dedekind-finite). But there are examples, similar in nature, but different from, the example I gave due to Cohen, that this weakened definition is not equivalent to being finite, or being Dedekind-finite.
