It is not. Using set forcing, we can add 'undefinable' reals in a controlled manner, while keeping complexity of parameter-free definable sets low.

Specifically, let $M$ be a countable $ω$-model of ZFC\P, real $r$ be Cohen generic over $M$, and $ℚ_M(r)$ be the minimal field of reals containing $r$ and all reals in $M$. Then, in $ℚ_M(r)$, every parameter-free definable bounded subset has the least upper bound, but $\sqrt{|r|}$ does not exist.

Because $r$ is transcendental over $M$, $\sqrt{|r|}∉ℚ_M(r)$.

Next, let $X$ be a bounded set of reals parameter-free definable in $ℚ_M(r)$. $ℚ_M(r)$ may not be closed under square roots, but it witnesses that the Cohen forcing is homogeneous, and therefore $\sup(X)∈M$, as required.

In more detail, by genericity, $\sup(X)$ can be determined with arbitrary given precision by a forcing condition (with the forcing relation definable in $M$). For Cohen forcing (modulo choice of representation), the conditions are rational $p<t$ (asserting $p<r<t$). Now, if $r$ is Cohen generic, then so is $qr$ for all nonzero $q∈ℚ$, and $ℚ_M(qr) = ℚ_M(r)$ (and same with $q+r$), and therefore all conditions lead to the same $\sup(X)$.

An interesting remaining question is whether there are computable examples.

It is not. Using set forcing, we can add 'undefinable' reals in a controlled manner, while keeping complexity of parameter-free definable sets low.

Specifically, let $M$ be a countable $ω$-model of ZFC\P, real $r$ be Cohen generic over $M$, and $ℝ_M(r)$ be the minimal field of reals containing $r$ and all reals in $M$ (in the initial revision, I called it $ℚ_M(r)$; it is a proper subset of $ℝ^{M[r]}$). Then, in $ℝ_M(r)$, every parameter-free definable bounded subset has the least upper bound, but $\sqrt{|r|}$ does not exist.

Because $r$ is transcendental over $M$, $\sqrt{|r|}∉ℝ_M(r)$.

Next, let $X$ be a bounded set of reals parameter-free definable in $ℝ_M(r)$. $ℝ_M(r)$ may not be closed under square roots, but it witnesses that the Cohen forcing is homogeneous, and therefore $\sup(X)∈M$, as required.

In more detail, by genericity, $\sup(X)$ can be determined with arbitrary given precision by a forcing condition (with the forcing relation definable in $M$). For Cohen forcing (modulo choice of representation), the conditions are rational $p<t$ (asserting $p<r<t$). Now, if $r$ is Cohen generic, then so is $qr$ for all nonzero $q∈ℚ$, and $ℝ_M(qr) = ℝ_M(r)$ (and same with $q+r$), and therefore all conditions lead to the same $\sup(X)$.

An interesting remaining question is whether there are computable examples.