Let $K$ be a global field and $\upsilon$ a place of $K$. Let $K_{\upsilon}$ denote the completion of $K$ at $\upsilon$ and $K_{(\upsilon)}:=K^{sep}\cap K_{\upsilon}$ the henselization (which is the quotient field of the henselization of the discrete valuation ring in case of a non-archimedean valuation).

Let $V$ be a $K_{(\upsilon)}$ variety with a point $\operatorname{Spec}(K_{\upsilon}) \to V$. I have seen used in a paper that in this case $V$ has to have a $K_{(\upsilon)}$-point. Why is this true?

I am thankful for any thoughts and ideas as I am not so familiar with algebraic geometry and maybe missed a result that would be helpful here.

Thoughts so far:

- If $\upsilon$ is complex, then $K_{(\upsilon)}$ is separably closed and the statement is true.
- If $\upsilon$ is non-archimedian, let $R^h$ denote the henselization of the corresponding discrete valuation ring and let $t$ be a generator of the maximal ideal. By a result of Greenberg a $R^h$-variety has a $R^h$-point iff it has a $R^h/t^n$-point for all $n \geq 1$. I fail to see how a point $\operatorname{Spec}(K_{\upsilon})=\operatorname{Spec}(\operatorname{Quot}(\varprojlim R^h/t^n)) \to V$ gives a point $\operatorname{Spec}(\varprojlim R^h/t^n) \to V$ (in the case that $V$ is not proper), such that the result can be used.
- If $\upsilon$ is real, I have no idea where to start.