Some very weak statements on choice

Question

This is a follow-up question to Does $|(X\times\{0\}) \cup (X\times\{1\})| \leq |X|$ for $X$ infinite imply ${\sf AC}$?

Consider the statements

$(\text{S}1)$ For any infinite set $X$ there is an injection $\varphi$ from $(X\cup\{X\})$ into $X$.

and

$(\text{S}2)$ For any infinite set $X$ there is an injection $\varphi$ from $(X\times\{0\}) \cup (X\times\{1\})$ into $X$.

Is it true that ${\sf ZF} < {\sf ZF}+(\text{S1}) < {\sf ZF}+(\text{S}2)$? (Here $<$ means "strictly weaker".)

Answer 1

$S1$ is equivalent to "every infinite set is Dedekind-infinite", so $\sf ZF<ZF+(S1)$, because it is consistent that infinite Dedekind-finite sets exist.

$\sf ZF+(S1)<ZF+(S2)$ follows using results from this answer, which implies that $\sf DC_\kappa$ doesn't imply $S2$, and hence nonexistence of infinite Dedekind-finite sets, which is a corollary to $\sf DC(=DC_\omega)$.