This question may be easy and indicative of my ignorance about the failure of the axiom of choice. If so, I apologize. Below assume $\mathsf{DC}$ but not $\mathsf{AC}$. Suppose we have a partial order $A = (A, \leq)$ satisfying the following:

$A$ is uncountable in the sense that there is no surjection of $\omega$ onto it.

$\leq$ is a transitive partial order so that for any $a \in A$ the set of $b \leq a$ is linearly ordered

$\leq$ is well founded: there is no infinite $ \leq$-descending sequence

There is no uncountable $\leq$-linearly ordered subset of $A$

For each $a \in A$, the set of $b\in A$ for which there is an order isomorphism between $\{c \; | \; c < b\}$ and $\{ c\; | \; c < a\}$ is countable.

Of course if choice holds then we call such an object an Aronszajn tree. What I want to know is whether, under simply $\mathsf{DC}$, this is still acceptable. Specifically, do the above conditions guarantee the existence of a rank function into $\omega_1$ so that we can make sense of "the $\alpha^{\rm th}$ level of $T$" ?

More to the point is the following. It's a well known theorem of Solovay that under $\mathsf{AD}$ $\omega_1$ is measurable and hence there are no Aronszajn trees under $\mathsf{AD}$. Does this suffice to rule out the existence of the partial order $A$ described above?