You can say a bit more than just every infinite set surjects onto $\omega$. Every infinite set is Dedekind infinite.

The reason is simple. If $T$ is an infinite transitive set, then $T\cap V_\omega$ is an infinite transitive set, and it is countable. To see this, note that if $T\cap V_\omega$ is finite, then, since $T$ is infinite, there is a gap in the finite ranks, but this is impossible as the rank function on a transitive set has a downwards closed image (i.e., its image is an ordinal).

This can probably be lifted to get slightly more out of this argument, but the "inspect locally, conclude globally" is bound to fail here. If every infinite set of reals is Dedekind-infinite, then every set of reals is equipotent to a transitive set. Simply consider the reals as $V_{\omega+1}$ and then every infinite set $T\subseteq V_{\omega+1}$ is equipotent with $T\cup V_\omega$, which is transitive. Equally, use $\mathcal P(\omega)$ and take $T\cup\omega$.

But, remember that we can have sets of reals which are very eccentric, in the technical sense, while still being Dedekind-infinite. So there is no reason to expect that the above argument can be used for anything significantly stronger.

You can say a bit more than just every infinite set surjects onto $\omega$. Every infinite set is Dedekind infinite.

The reason is simple. If $T$ is an infinite transitive set, then $T\cap V_\omega$ is an infinite transitive set, and it is countable. To see this, note that if $T\cap V_\omega$ is finite, then, since $T$ is infinite, there is a gap in the finite ranks, but this is impossible as the rank function on a transitive set has a downwards closed image (i.e., its image is an ordinal).

This can probably be lifted to get slightly more out of this argument, but the "inspect locally, conclude globally" is bound to fail here. If every infinite set of reals is Dedekind-infinite, then every set of reals is equipotent to a transitive set. Simply consider the reals as $V_{\omega+1}$ and then every infinite set $T\subseteq V_{\omega+1}$ is equipotent with $T\cup V_\omega$, which is transitive. Equally, use $\mathcal P(\omega)$ and take $T\cup\omega$.

But, remember that we can have sets of reals which are very eccentric, in the technical sense, while still being Dedekind-infinite. So there is no reason to expect that the above argument can be used for anything significantly stronger.

Here's a nice thing to ponder about while walking in a park.

Assume $\sf KWP_1$ and for a set $x$ let $\kappa_x$ be the least such that $x$ maps into $\mathcal P(\kappa_x)$. Assume the following holds: $\kappa_x<\aleph(x)$ for all $x$. Then $\sf TC$ holds.

To see that, simply replace $x$ with a copy in $\mathcal P(\kappa_x)$, and since $\kappa_x+x=x$, as it is Dedekind-infinite, we can assume that said copy also contains $\kappa_x$ itself.

So, firstly, this seems to be flexible enough to extend to $\sf KWP_\alpha$, with some obvious caveats that we need to have more conditions on how the copy behaves with respect to adding subsets of a lower Kinna–Wagner rank. This might end up a bit too messy.

Secondly, is this consistent without choice at all? It seems likely, presumably if we violate choice on the reals in a way that preserves $\sf KWP_1$, this should work out. But I don't have a proof off the top of my head.

You can say a bit more than just every set surjects onto $\omega$. Every infinite set is Dedekind infinite.

The reason is simple. If $T$ is an infinite transitive set, then $T\cap V_\omega$ is an infinite transitive set, and it is countable. To see this, note that if $T\cap V_\omega$ is finite, then, since $T$ is infinite, there is a gap in the finite ranks, but this is impossible as the rank function on a transitive set has a downwards closed image (i.e., its image is an ordinal).

This can probably be lifted to get slightly more out of this argument, but the "inspect locally, conclude globally" is bound to fail here. If every infinite set of reals is Dedekind-infinite, then every set of reals is equipotent to a transitive set. Simply consider the reals as $V_{\omega+1}$ and then every infinite set $T\subseteq V_{\omega+1}$ is equipotent with $T\cup V_\omega$, which is transitive. Equally, use $\mathcal P(\omega)$ and take $T\cup\omega$.

But, remember that we can have sets of reals which are very eccentric, in the technical sense, while still being Dedekind-infinite. So there is no reason to expect that the above argument can be used for anything significantly stronger.

You can say a bit more than just every infinite set surjects onto $\omega$. Every infinite set is Dedekind infinite.

The reason is simple. If $T$ is an infinite transitive set, then $T\cap V_\omega$ is an infinite transitive set, and it is countable. To see this, note that if $T\cap V_\omega$ is finite, then, since $T$ is infinite, there is a gap in the finite ranks, but this is impossible as the rank function on a transitive set has a downwards closed image (i.e., its image is an ordinal).

This can probably be lifted to get slightly more out of this argument, but the "inspect locally, conclude globally" is bound to fail here. If every infinite set of reals is Dedekind-infinite, then every set of reals is equipotent to a transitive set. Simply consider the reals as $V_{\omega+1}$ and then every infinite set $T\subseteq V_{\omega+1}$ is equipotent with $T\cup V_\omega$, which is transitive. Equally, use $\mathcal P(\omega)$ and take $T\cup\omega$.

But, remember that we can have sets of reals which are very eccentric, in the technical sense, while still being Dedekind-infinite. So there is no reason to expect that the above argument can be used for anything significantly stronger.