Let $D$ be a domain with quotient field $K$.
The ring of integer valued polynomials is defined as
$$
\operatorname{Int}(D) = \{\, f \in K[X] \mid f(D) \subset D \,\}.
$$
It is known that if $D$ is a BF-domain, then so is $\operatorname{Int}(D)$.
If $D$ is a Dedekind domain with finite residue fields, then $\operatorname{Int}(D)$ is a Prüfer domain (in particular integrally closed), but does not satisfy the ascending chain condition on divisorial ideals. Hence it is not a Krull domain.
Thus, for instance, the rather beautiful ring $\operatorname{Int}(\mathbb Z)$ is a non-Krull integrally closed BF-domain.
See the book Integer Valued Polynomials by Cahen and Chabert, in particular Chapter VI, for background on integer valued polynomials. (But note that some of the results on factorization properties are outdated.)
An example which is moreover not Prüfer:
This example can be adapted to obtain a ring which is (completely) integrally closed, not Krull, not Prüfer, and BF. Let me first recall how a bunch of ring-theoretic properties behave for $\operatorname{Int}(D)$.
If $\operatorname{Int}(D)$ is Prüfer, then $D$ is an almost Dedekind domain (all localizations at maximal, equivalently prime, ideals are discrete valuation rings) with all residue fields finite. In particular, almost Dedekind domains are $1$-dimensional Prüfer domains.
Moreover, $\operatorname{Int}(D)$ is integrally closed if and only if $D$ is. Similarly, $\operatorname{Int}(D)$ is completely integrally closed (c.i.c.) if and only if $D$ is. (For Noetherian domains, being c.i.c. and being integrally closed are equivalent. In the non-Noetherian case, being c.i.c is a stronger property.)
Recall a convenient ring-theoretic characterization of Krull domains: A domain $D$ is Krull if and only if it is completely integrally closed and a Mori domain (satisfies the ascending chain condition on divisorial ideals).
It is also known (see P.-J. Cahen, S. Gabelli, E. Houston, Mori domains of integer-valued polynomials, J. Pure. Appl. Algebra, 153(1), 2000) that $\operatorname{Int}(D)$ Mori implies $D$ Mori.
So, to get $\operatorname{Int}(D)$ to be BF, integrally closed, non-Prüfer, non-Krull we want that $D$ is
- BF,
- integrally closed,
- not an almost Dedekind domain (to ensure non-Prüfer),
- not c.i.c. or not Mori (to ensure non-Krull).
Actually, it would be better if $D$ is even c.i.c. but not Mori.
Here are two possibilities:
Take $D=K[X,XY,XY^2,XY^3,\ldots] \subset K[X,Y]$. By exercise I.3.14 in L. Fuchs, L. Salce, Modules Over Non-Noetherian Domains, $D$ is integrally closed but not c.i.c. It is easy to see that it is 2-dimensional and BF.
Incidentally, $D=\operatorname{Int}(\mathbb Z)$ is c.i.c., non-Krull, 2-dimensional and BF. So
$$
\operatorname{Int}(\operatorname{Int}(\mathbb Z))
$$
is non-Krull, non-Prüfer, c.i.c., BF. I haven't seen this ring before; it's rather amusing that this works! You can in fact replace $\mathbb Z$ by any Dedekind domain with finite residue fields and get the same result.
Edit: The following may be an easier example. If $D$ is integrally closed, non-Krull, BF then so is $D[X]$. ($D[X]$ is Krull iff $D$ is Krull.) $D[X]$ can moreover not be Prüfer ($D[X]$ is Prüfer iff $D$ is a field). So, for instance $\operatorname{Int}(\mathbb Z)[X]$ also works. Unfortunately, I don't know enough about integer valued polynomials to see quickly whether actually $\operatorname{Int}(\mathbb Z)[X]$ is a proper subring of $\operatorname{Int}(\operatorname{Int}(\mathbb Z))$ or whether they are equal. (If a domain $D$ is an intersection of localizations at prime ideals with infinite residue fields, then $\operatorname{Int}(D)=D[X]$ is trivial.)
