This is a non-answer too long for a comment.

If $K$ is an ordered field that satisfies the least upper bound property for sets definable without parameters, then suprema of sets definable with parameters can be arbitrarily closely approximated in $K$.

Consequently, the completion of $K$ is a real-closed field (and $K$ itself is a real-closed field if it happens to be henselian).

To see this, let $X=\{x:K\models\phi(x,\vec a)\}$ be a nonempty bounded set. Applying an affine transformation if necessary, we may assume $0\in X\subseteq(-\infty,1)$. Let $\psi(w)$ denote the formula $$\begin{multline*} e>0\land\forall\vec u\:\bigl[\phi(0,\vec u)\land\forall x\:(\phi(x,\vec u)\to x<1)\\\to\exists z\:\bigl(\phi(z,\vec u)\land\forall x\:(\phi(x,\vec u)\to x<z+e)\bigr)\bigr]. \end{multline*}$$ In words, $\psi(e)$ says that for every set of the form $Y=\{x:K\models\phi(x,\vec u)\}$ for some $\vec u\in K$, if $0\in Y\subseteq(-\infty,1)$, then $\sup Y$ can be approximated within distance $e$, in the sense that there exists $z\in Y$ such that $Y\subseteq(-\infty,z+e)$.

Now, $E=\{e:K\models\psi(e)\}$ is definable without parameters, and clearly $1\in E\subseteq(0,+\infty)$, hence by assumption (or rather, by the equivalent greatest lower bound property), there exists $e=\inf E\ge0$.

If $e=0$, we are done. However, $e>0$ is impossible, because $$e\in E\implies e/2\in E.$$ Indeed, using the notation above, if $z\in Y\subseteq(-\infty,z+e)$, then either $z\in Y\subseteq(-\infty,z+e/2)$, or there exists $z'\ge z+e/2$ such that $z'\in Y\subseteq(-\infty,z'+e/2]$.

This is a non-answer too long for a comment.

If $K$ is an ordered field that satisfies the least upper bound property for sets definable without parameters, then suprema of sets definable with parameters can be arbitrarily closely approximated in $K$.

Consequently, the completion of $K$ is a real-closed field (and $K$ itself is a real-closed field if it happens to be henselian).

To see this, let $X=\{x:K\models\phi(x,\vec a)\}$ be a nonempty bounded set. Applying an affine transformation if necessary, we may assume $0\in X\subseteq(-\infty,1)$. Let $\psi(w)$ denote the formula $$\begin{multline*} e>0\land\forall\vec u\:\bigl[\phi(0,\vec u)\land\forall x\:(\phi(x,\vec u)\to x<1)\\\to\exists z\:\bigl(\phi(z,\vec u)\land\forall x\:(\phi(x,\vec u)\to x<z+e)\bigr)\bigr]. \end{multline*}$$ In words, $\psi(e)$ says that for every set of the form $Y=\{x:K\models\phi(x,\vec u)\}$ for some $\vec u\in K$, if $0\in Y\subseteq(-\infty,1)$, then $\sup Y$ can be approximated within distance $e$, in the sense that there exists $z\in Y$ such that $Y\subseteq(-\infty,z+e)$.

Now, $E=\{e:K\models\psi(e)\}$ is definable without parameters, and clearly $1\in E\subseteq(0,+\infty)$, hence by assumption (or rather, by the equivalent greatest lower bound property), there exists $e_0=\inf E\ge0$.

If $e_0=0$, we are done. However, $e_0>0$ is impossible, because $$e\in E\implies e/2\in E.$$ Indeed, using the notation above, if $z\in Y\subseteq(-\infty,z+e)$, then either $z\in Y\subseteq(-\infty,z+e/2)$, or there exists $z'\ge z+e/2$ such that $z'\in Y\subseteq(-\infty,z'+e/2]$.