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Edit: As Emil Jeřábek has pointed out in the comments, there is an error in the paper linked below, and the example does not work. I'll leave the answer here to document the error and add to the interesting list of attempted answers gathered here.

Let me add that Dmytro Taranovsky has given a correct example, but I would still be very interested to see a "forcing-free" and especially an "algebraically natural" example of a $\emptyset$-definably complete ordered field which is not definably complete.

This question was answered negatively by Mojtaba Moniri in his recent paper On definable completeness for ordered fields, Reports on Mathematical Logic, Number 54 (2019), pp. 95-100.

An explicit example is given: $\mathbb{R}((t^\Gamma))$, where $\Gamma$ is the ordered additive group of dyadic rationals. Here $\mathbb{R}((t^\Gamma))$ is the Hahn series field over $\mathbb{R}$ with value group $\Gamma$: its elements are formal sums $\sum_{e\in \Gamma}c_et^e$ with $c_e\in \mathbb{R}$ such that the support $\{e\in \Gamma\mid c_e\neq 0\}$ is well-ordered. This field is complete for cuts definable without parameters, but not real closed: $t$ has no cube root.

Edit: As Emil Jeřábek has pointed out in the comments, there is an error in the paper linked below, and the example does not work. I'll leave the answer here to document the error and add to the interesting list of attempted answers gathered here. An erratum has now appeared: Corrigendum to "[..]" RML, referencing this discussion.

Let me add that Dmytro Taranovsky has given a correct example, but I would still be very interested to see a "forcing-free" and especially an "algebraically natural" example of a $\emptyset$-definably complete ordered field which is not definably complete.

This question was answered negatively by Mojtaba Moniri in his recent paper On definable completeness for ordered fields, Reports on Mathematical Logic, Number 54 (2019), pp. 95-100.

An explicit example is given: $\mathbb{R}((t^\Gamma))$, where $\Gamma$ is the ordered additive group of dyadic rationals. Here $\mathbb{R}((t^\Gamma))$ is the Hahn series field over $\mathbb{R}$ with value group $\Gamma$: its elements are formal sums $\sum_{e\in \Gamma}c_et^e$ with $c_e\in \mathbb{R}$ such that the support $\{e\in \Gamma\mid c_e\neq 0\}$ is well-ordered. This field is complete for cuts definable without parameters, but not real closed: $t$ has no cube root.

This question was answered negatively by Mojtaba Moniri in his recent paper On definable completeness for ordered fields, Reports on Mathematical Logic, Number 54 (2019), pp. 95-100.

An explicit example is given: $\mathbb{R}((t^\Gamma))$, where $\Gamma$ is the ordered additive group of dyadic rationals. Here $\mathbb{R}((t^\Gamma))$ is the Hahn series field over $\mathbb{R}$ with value group $\Gamma$: its elements are formal sums $\sum_{e\in \Gamma}c_et^e$ with $c_e\in \mathbb{R}$ such that the support $\{e\in \Gamma\mid c_e\neq 0\}$ is well-ordered. This field is complete for cuts definable without parameters, but not real closed: $t$ has no cube root.

Edit: As Emil Jeřábek has pointed out in the comments, there is an error in the paper linked below, and the example does not work. I'll leave the answer here to document the error and add to the interesting list of attempted answers gathered here.

Let me add that Dmytro Taranovsky has given a correct example, but I would still be very interested to see a "forcing-free" and especially an "algebraically natural" example of a $\emptyset$-definably complete ordered field which is not definably complete.

This question was answered negatively by Mojtaba Moniri in his recent paper On definable completeness for ordered fields, Reports on Mathematical Logic, Number 54 (2019), pp. 95-100.

An explicit example is given: $\mathbb{R}((t^\Gamma))$, where $\Gamma$ is the ordered additive group of dyadic rationals. Here $\mathbb{R}((t^\Gamma))$ is the Hahn series field over $\mathbb{R}$ with value group $\Gamma$: its elements are formal sums $\sum_{e\in \Gamma}c_et^e$ with $c_e\in \mathbb{R}$ such that the support $\{e\in \Gamma\mid c_e\neq 0\}$ is well-ordered. This field is complete for cuts definable without parameters, but not real closed: $t$ has no cube root.

This question was answered negatively by Mojtaba Moniri in his recent paper On definable completeness for ordered fields, Reports on Mathematical Logic, Number 54 (2019), pp. 95-100.

An explicit example is given: $\mathbb{R}((t^\Gamma))$, where $\Gamma$ is the ordered additive group of dyadic rationals. Here $\mathbb{R}((t^\Gamma))$ is the Hahn series field over $\mathbb{R}$ with value group $\Gamma$: its elements are formal sums $\sum_{e\in \Gamma}c_et^e$ such that the support $\{e\in \Gamma\mid c_e\neq 0\}$ is well-ordered. This field is complete for cuts definable without parameters, but not real closed: $t$ has no cube root.

This question was answered negatively by Mojtaba Moniri in his recent paper On definable completeness for ordered fields, Reports on Mathematical Logic, Number 54 (2019), pp. 95-100.

An explicit example is given: $\mathbb{R}((t^\Gamma))$, where $\Gamma$ is the ordered additive group of dyadic rationals. Here $\mathbb{R}((t^\Gamma))$ is the Hahn series field over $\mathbb{R}$ with value group $\Gamma$: its elements are formal sums $\sum_{e\in \Gamma}c_et^e$ with $c_e\in \mathbb{R}$ such that the support $\{e\in \Gamma\mid c_e\neq 0\}$ is well-ordered. This field is complete for cuts definable without parameters, but not real closed: $t$ has no cube root.