dual Dedekind-infinity may not imply Dedekind-infinite without AC It is written in wikipedia: 
https://en.wikipedia.org/wiki/Dedekind-infinite_set
It is not provable (in ZF without the AC) that dual Dedekind-infinity implies that A is Dedekind-infinite. (For example, if B is an infinite but Dedekind-finite set, and A is the set of finite one-to-one sequences from B, then "drop the last element" is a surjective but not injective function from A to A, yet A is Dedekind finite.)
A is set of all finite subsets of B. Hence it has to be infinite and of same cardinality as of A. But 'drop the last elemenet' but which element.(For this we need choosing map from [0,n] to elements of A but then it requires AC but then these two definitions become equivalent.)
And why A is Dedekind finite.
 A: You have two mistakes, as pointed by Andres in the comments.

*

*The claim that for an infinite set $A$, the set $B=\{X\subseteq A\mid X\text{ finite}\}$ has the same cardinality as $A$ requires the axiom of choice. For a Dedekind-finite set this is most certainly false in any case (there's an injection from $A$ into $B$, and this injection is not a surjection. So if $|A|=|B|$, it is impossible that $B$ is Dedekind-finite, so $A$ is not either.)


*Sets have no linear ordering in most cases. So there's no real meaning to "last element". You can talk about the set of finite sequences, but those are always Dedekind-infinite, since $\{\langle a\rangle,\langle a,a\rangle,\langle a,a,a\rangle,\ldots\}$ is a countably infinite subset.
However, we can show that the set of injective finite sequences is in fact Dedekind-finite, when $A$ is Dedekind-finite.
Now we return to your question, if $A$ is Dedekind-finite, and $S$ is the set of injective finite sequences of $A$, then $B$ is Dedekind-finite, and the operation "remove the last element" is well-defined, because the elements of $B$ are sequences, and no choice of ordering is needed.
