I have trouble understanding why $$\Diamond P \to \square\Diamond P$$ is valid in Euclidean frames.
I found a proof online which is detailed as follows:
Proof. Suppose $F$ is a Euclidean frame, and $M$ a model based on $F.$ Suppose $\models_w \Diamond A$. Then there is a $v$ such that $w\mathrel{R}v$ and $\models_v A.$ Now, for any $u$ with $w\mathrel{R}u$, we have $u\mathrel{R}v$ since $R$ is Euclidean. So $\models_u \Diamond$A. Since $u$ is arbitrary, $\models_w \Box\Diamond A,$ and therefore $\models_w \Diamond A \to \Box\Diamond A.$
My question is what if the arbitrary $u$ is chosen as world $v.$ My understanding of necessarily true is that from every world from the current world should satisfy the condition. However, at world $v,$ there is no path to another world where $p$ is true since $p$ is only true at $v$ itself. Doesn't that imply that $\Diamond A$ is false at $v$ and hence $\Box\Diamond A$ is false as well? If that's the case, why is the axiom valid?
Thank you in advance for any explanation!