False: Prikry forcing is ccc

For each $\alpha <\kappa$, set $A_{\alpha }=\{\xi <\kappa :\alpha <\xi \}$. Its complement is the intersection-complement of the singletons $\{\xi \}$ for $\xi \le \alpha$: nonprincipality puts every $\kappa \setminus \{\xi \}$ in $U$, and F3 keeps their fewer-than-$\kappa$ intersection $A_{\alpha }$ in $U$.

Define $p_{\alpha }=(⟨\alpha ⟩,A_{\alpha })$. Since $\min A_{\alpha }=\alpha +1$, F2 makes $p_{\alpha }$ a condition. If $\alpha \mathrm{\ne }\beta$ and $r$ extended both $p_{\alpha }$ and $p_{\beta }$, the stem of $r$ would end-extend both one-entry stems, so its first entry would have to equal both $\alpha$ and $\beta$. This is impossible. Therefore $\{p_{\alpha }:\alpha <\kappa \}$ is a pairwise incompatible family, and the indexing is injective, so it is an antichain of cardinality exactly $\kappa$.

Because $\kappa$ is uncountable, the antichain in step 2.1 is uncountable. By F4 this violates ccc, furnishing the promised witness to the failure of the refuted statement.

There is no conflict with the weaker positive conclusion in F1. There are only $\kappa$ finite stems, and conditions with the same stem are compatible by F2. Thus among ${\kappa }^{+}$ conditions two share a stem and are compatible, so no antichain has size ${\kappa }^{+}$. The explicit antichain from step 2.1 has size $\kappa$, which is below that forbidden size.