The implication $ACC \implies UCC$ is irreversible in $ZF$. This follows again from the transfer theorem of Pincus:
1) $UCC$ is an injectively boundable statement, see note 103 in "Consequences of the axiom of choice" by Howard & Rubin. Right after the theorem in pp. 285 and its corollary, there are examples of statements of this kind, one of which is form 31 (which is precisely $UCC$). The fact that this is the case follows in turn from the application of lemma 3.5 in Howard, P.-Solski, J.: "The strength of the $\Delta$-system lemma", Notre Dame J. Formal Logic vol 34, pp. 100-106 - 1993
2) It is known that $¬ACC$ is boundable, and hence injectively boundable.
Then we can apply the transfer theorem of Pincus to the conjunction $UCC \wedge ¬ACC$ and we are done.
SECOND PROOF: Browsing through the "Consequences..." book I've just found another less direct proof of the same fact. I'm adding it here to avoid the interested person from looking it up for himself (especially since the AC website is not working these days). It involves form 9, known as $W_{\aleph_0}$: a set is finite if and only if it is Dedekind finite. Now, $ACC \implies W_{\aleph_0}$ is provable in $ZF$ (in fact this was already proved by Dedekind), while $UCC$ does not imply $W_{\aleph_0}$. The latter follows from the fact that in the basic Fraenkel model, $\mathcal{N}1$, $ACC$ UCC$is valid while$W_{\aleph_0}$is not, and such a result is transferable by considerations of Pincus that can be found at Pincus, D.: "Zermelo-Fraenkel consistency results by Fraenkel Mostowski methods", J. of Symbolic Logic vol 37, pp. 721-743 - 1972 2 added 944 characters in body The implication$ACC \implies UCC$is irreversible in ZF.$ZF$. This follows again from the transfer theorem of Pincus: 1)$UCC$is an injectively boundable statement, see note 103 in "Consequences of the axiom of choice" by Howard & Rubin. Right after the theorem in pp. 285 and its corollary, there are examples of statements of this kind, one of which is form 31 (which is precisely$UCC$). The fact that this is the case follows in turn from the application of lemma 3.5 in Howard, P.-Solski, J.: "The strength of the$\Delta$-system lemma", Notre Dame J. Formal Logic vol 34, pp. 100-106 - 1993 2) It is known that$¬ACC$is boundable, and hence injectively boundable. Then we can apply the transfer theorem of Pincus to the conjunction$UCC \wedge ¬ACC$and we are done. SECOND PROOF: Browsing through the "Consequences..." book I've just found another less direct proof of the same fact. I'm adding it here to avoid the interested person from looking it up for himself (especially since the website is not working these days). It involves form 9, known as$W_{\aleph_0}$: a set is finite if and only if it is Dedekind finite. Now,$ACC \implies W_{\aleph_0}$is provable in$ZF$(in fact this was already proved by Dedekind), while$UCC$does not imply$W_{\aleph_0}$. The latter follows from the fact that in the basic Fraenkel model,$\mathcal{N}1$,$ACC$is valid while$W_{\aleph_0}$is not, and such a result is transferable by considerations of Pincus that can be found at Pincus, D.: "Zermelo-Fraenkel consistency results by Fraenkel Mostowski methods", J. of Symbolic Logic vol 37, pp. 721-743 - 1972 1 The implication$ACC \implies UCC$is irreversible in ZF. This follows again from the transfer theorem of Pincus: 1)$UCC$is an injectively boundable statement, see note 103 in "Consequences of the axiom of choice" by Howard & Rubin. Right after the theorem in pp. 285 and its corollary, there are examples of statements of this kind, one of which is form 31 (which is precisely$UCC$). The fact that this is the case follows in turn from the application of lemma 3.5 in Howard, P.-Solski, J.: "The strength of the$\Delta$-system lemma", Notre Dame J. Formal Logic vol 34, pp. 100-106 - 1993 2) It is known that$¬ACC$is boundable, and hence injectively boundable. Then we can apply the transfer theorem of Pincus to the conjunction$UCC \wedge ¬ACC\$ and we are done.