TITLE: Trivialisation of vector bundles on Stein spaces
QUESTION [10 upvotes]: Does every vector bundle on a Stein space have a finite local trivialisation?
Definitions:

Stein space means either a complex analytic Stein space or a nonarchimedean Stein space in the sense of Kiehl.
Vector bundle means either holomorphic vector bundle or rigid analytic vector bundle (= locally free sheaf). Edit: Say, of constant rank $r$.
finite local trivialisation = local trivialisation where the covering consists of finitely many sets.

In the nonarchimedean case, assume moreover that the base field is spherically complete or that the Stein space is unbounded.
My thoughts so far: Usually, the space is noncompact, so the answer is probably "no". On the other hand, I could not come up with an example yet. If the Stein space is the analytification of an algebraic variety, then at least those vector bundles that are analytifications of algebraic bundles will have a finite trivialisation. Every vector bundle on a Stein space is determined by finitely many global sections. Could that help? For an example, one should probably use the non-finitely generated ideals in the algebra of global functions of the Stein space. I cannot write down any of them either.
As said, I suspect the answer to be "no". It would be great if somebody knew an example or a reference.

REPLY [4 votes]: I think that it is also possible to give a positive answer to the question using purely topological arguments, namely Ostrand's theorem: for every paracompact space $X$ of dimension $n$ and every open covering $\mathscr U$ of $X$, there exists $n+1$ families $\mathscr V_1,\dotsc, \mathscr V_{n+1}$ of disjoint open sets whose union is a covering that refines $\mathscr U$. 
Apply this to a covering trivializing your vector bundle $E$. Then, for every $i$, $E$ is trivial on the open set that is the union of the elements of $\mathscr V_i$, and you are done.
Note that the argument does use the fact that the space is finite-dimensional, but not so much the Stein hypothesis (maybe for paracompactness).