Interesting question! I haven't checked it yet, but myMy guess would be that just using LEM a relation is well-founded iff it does not contain an entire inverse subrelation. Turning this into a descending sequence represented as function from the natural numbers then sounds like dependend choice maybe?
Edit: I have checked the argument now in Coq and while the answers given in the meantime perfectly explain the equivalence to DC, here's a brief addition/refinement regarding the use of DC depending on the concrete formulation of infinite descent.
The usual definition Théo already suggested by the notation $x_0>x_1>x_2>\dots$ is based on a function $f:\mathbb N \to X$ representing a descending chain. However, descent can also be formulated as a non-empty set $P$ such that for all $x\in P$ there is $y\in P$ with $y< x$. This notion of an entire subrelation shows up in Daniel's answer and its absence is indeed equivalent to well-foundedness just using LEM:
If $<$ is well-founded, then there is no entire subrelation $P$. Proof: As before by induction on Acc for the witness $x\in P$ of non-emptiness.
Given LEM, if there is no entire subrelation, then $<$ is well-founded. Proof: Exactly as before, suppose $x$ were not accessible, then the inaccessible points form an entire subrelation.
This now localises the use of DC, since this is exactly the requirement to turn an entire subrelation into a chain, the other direction is of course for free:
If there is no entire subrelation, then there is no descending chain. Proof: Assume $f$ were a descending chain, then its range induces an entire subrelation.
Given DC, if there is no descending chain, then there is no entire subrelation. Proof: Assume $P$ were entire, then using DC on the type $\Sigma P$ and the relation $\lambda x y.\, y < x$ yields a descending chain.
(Of course the above implications are actually proven in the stronger contrapostive form but I chose the weaker form to align with well-foundedness.)
