$\DeclareMathOperator\Spa{Spa}$Fargues's Theorem for $\Spa(C,O_C)$ states that the category of (mixed characteristic) shtukas with one paw at $x_C$ is equivalent to the category of Breuil-Kisin-Fargues modules over $A_{inf}=W(O_C)$. (Here, $C$ is an algebraically closed perfectoid field, and $O_C$ is the ring of integers of $C$.)

My question is concerned with extending this theorem to shtukas over $\Spa(C,C^+)$, where $C^+$ is an open and bounded valuation subring of $C$ and BKF-modules over $W(C^+)=A_{inf}(C,C^+)$ (and $C$ is still an algebraically closed perfectoid field).

First question: Where in the proof of Fargues's theorem does this slighly bigger generality fail?

A sketch of the proof of Fargues's theorem goes as follows:

Step 1: Turn a shtuka over $\mathcal{Y}_{[0,\infty)}$ to a shtuka over $\mathcal{Y}_{[0,\infty]}$. This step is done by glueing the shtuka with a "spread" vector bundle over the Fargues–Fontaine (that is the corresponding $\phi$-module over $\mathcal{Y}_{[r,\infty]}$ for some $r$). Here we use that we understand vector bundles over $\mathrm{FF}_{(C,O_C)}$ to be able to extend over $\infty$. Now theorem 8.7.7 in Kedlaya-Liu ("Relative $p$-adic Hodge Theory: Foundations" (MSN,arxiv)), tell us that vector bundles on $\mathrm{FF}_{(C,C^+)}$ coincide with those of $\mathrm{FF}_{(C,O_C)}$. That's why I think this step can be done as well.

Step 2: Turning an "analytic" shtuka over $\mathcal{Y}_{[0,\infty]}$ into an "algebraic" shtuka over $Y_{[0,\infty]}$. This is the content of Kedlaya's theorem 4.5.10 a) in his AWS (revised) notes, and that step is stated in full generality for any perfectoid Huber pair $(R,R^+)$.

Step 3: Turning the "algebraic" shtuka over $\operatorname{Spec}(A_{inf})\setminus \{\varpi=0=p\}$ into a BKF-module over $\operatorname{Spec}(A_{inf})$. This is now Kedlaya's theorem 4.5.10 b) of his AWS notes again. It works for any perfectoid field $K$ regardless of the chosen $K^+$.

All steps seem to work to me for $\Spa(C,C^+)$, but I might be missing a subtlety and I would like to know what that is. Step 1 is the one I would have most doubts about.