Let $X=\operatorname{Sp}(A)$ be an affinoid $K$-space, where $K$ is a p-adic field. Suppose that $X$ *lies in the interior* of another affinoid $K$-space $X'=\operatorname{Sp}(B)$. Recall that this means $X$ is an affinoid subdomain of $X'$ and $B$ has a system of affinoid generators $f_1,..., f_n$ over $K$ such that

$$X \subset \{ y \in X': \vert {f_i(y)} \vert <1 \}.$$

Now, let $U$ be an affinoid subdomain of $X'$ such that $X \subset \subset_{X'} U$, where this notation means that $U$ is *a wide open neighbourhood* of $X$ in $X'$ in the sense of Exercise 7.1.12 in Fresnel and van der Put's book on rigid analytic geometry. I believe that if $f,g \in \mathcal{O}_{X'}(U)$ are such that $f \vert_X = g \vert_X$ then there should exist an affinoid subdomain $W \subset U$ in $X'$ with $X \subset \subset_{X'} W$ such that $f \vert_W = g \vert_W$.

Irritatingly, I cannot prove this. All my proof attempts have been via contradiction because I cannot see any way to construct $W$ explictly. I have attempted to use characterisations of relative compactness in the associated Berkovich space $\mathcal{M}(X)$ of $X$ in order to reduce the problem to a topological one. I also thought that viewing $U$ as a strict neighbourhood of $X$ in $X'$ might be useful (see Fresnel and van der Put, Exercises 7.1.12, 7.7.1). However, neither of these approaches have allowed me to prove the result, so I am asking for help.

If anybody can point me to a proof of this result, or a counter-example, I would be very grateful. Thank you very much in advance - this has really been bugging me!