If $\xi$ is unramified at the valuation $v$, does the associated homogeneous space $C/K$ have a $K_v$-rational point?

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I'm looking at the following Corollary/Remark $(4.4-4.5)$ in Chapter X of Silverman's Arithmetic of Elliptic Curves:

Let $\phi:E/K\to E'/K$ be an isogeny defined over $K$, and $S$ a finite set of places (plus some conditions on $S$ which don't matter for my question). If $H^1(G_{\overline K/K},E[\phi];S)$ denotes the subset of $H^1(G_{\overline K/K},E[\phi])$ consisting of elements unramified outside of $S$, then $$S^{(\phi)}(E/K)\subseteq H^1(G_{\overline K/K},E[\phi],S).$$

Here, $S^{(\phi)}(E/K)$ is the Selmer group of $\phi$, which is equal to

$$S^{(\phi)}(E/K)=\ker\left(H^1(G_{\overline K/K},E[\phi])\to\prod_{v\in M_K}WC(E/K_v)\right),$$

where $M_K$ denotes the set of places on $K$, and $WC(E/K_v)$ is the Weil-Chatelet group of $E/K_v$.

We know at this point that for $\xi\in H^1(G_{\overline K/K},E[\phi])$, the image of $\xi$ in $WC(E/K_v)$ is nontrivial if and only if there exists an $K_v$-rational point on $C$, where $C$ is the homogeneous space associated to $\xi$. The author makes the following remark:

To determine whether a given element $\xi\in H^1(G_{\overline K/K},E[\phi],S)$ is in $S^{(\phi)}(E/K)$, one takes the corresponding homogeneous space $C/K$ for $\xi$ and checks whether $C(K_v)\neq\varnothing$ for the finitely many $v\in S$.

By definition and what we noted a paragraph above, it's true that $\xi$ is in $S^{(\phi)}(E/K)$ if and only if $C(K_v)\neq\varnothing$ for all valuations $v\in M_K$. However, by the remark above it seems that we only need to check the valuations $v$ where $\xi$ might be ramified? Does this mean then that the author is implicitly using the fact that if $\xi$ is unramified at $v$, then $C(K_v)\neq\varnothing$?

If this is the case, then how would I show this explicitly?

I've thought about it some but haven't gotten far.. if $\xi$ is unramified at $v$ then by definition, the image of $\xi$ under the "restriction" map $H^1(G_{\overline K/K},E[\phi])\to H^1(I_v,E[\phi])$ is zero, where $I_v$ is the inertia group of $v$. Or in other words, there is some $P\in E[\phi]$ such that $\xi_{\sigma}=P^{\sigma}-P$ for all $\sigma\in I_v$.

By basic properties of homogeneous spaces, we know that there exists an isomorphism $\theta:E\to C$ defined over $\overline K$. If we could show that $\theta$ is in fact defined over $K_v$, then we'd have $\theta(O)\in C(K_v)$. To do this, it suffices to show (I think?) that $\theta(Q^{\sigma})=\theta(Q)^{\sigma}$ for all $Q\in E$ and $\sigma\in G_{\overline K/K}$ which fix all elements of $K_v\cap\overline K$.

I'm not sure how to put these things together. How can I use the unramified assumption on $\xi$ to show that $\theta$ is defined over $K_v$? It seems that because we've chosen $\theta$ arbitrarily, we can't even guarantee it'll be defined over $K_v$. So maybe I need to explicitly construct $\theta$, but how would I do this? Any help is appreciated.