Let $k$ be a complete non-archimedean field and let $\varphi \colon X \to Y$ be a morphism of rigid analytic spaces over $k$, where $\newcommand{\Sp}{\operatorname{Sp}}Y = \Sp(B)$ is affinoid. Consider the following condition:

$(\dagger)$ The morphism $\varphi$ is separated and there exist two finite admissable affinoid coverings $\mathfrak{U} = (U_i)_{i \in I}$, $\mathfrak{V} = (V_i)_{i \in I}$ of $X$ such that $V_i \Subset_Y U_i$ (i.e. $V_i$ lies relatively compact in $U_i$ w.r.t. $Y$) for all $i \in I$.

A morphism $\varphi \colon X \to Y$ (with $Y$ not necessarily affinoid) is called proper if it is separated and there is an admissable affinoid covering $(W_j)_{j \in J}$ of $Y$ such that for each $j \in J$ the morphism $\varphi^{-1}(W_j) \to W_j$ satisfies condition $(\dagger)$.

If $Y$ is affinoid, then obviously a morphism satisfying condition $(\dagger)$ is proper, whereas the converse is not clear. In fact, Bosch in his book Lectures of Formal and Rigid Geometry writes at the beginning of section 6.4 that $(\dagger)$ is slightly stronger than properness.

Is there an example of a proper morphism with affinoid base which does not satisfy condition $(\dagger)$?