For some reason, I'm convinced the answer to the following question should be (obviously) negative, but I can't come up with a good reason.
Does there exist a number field $K$, a smooth projective geometrically connected curve $X/K$ of genus $g\geq 2$ with a $K$-rational point $x$ such that, for any number field $L/K$, $x$ does not intersect $ X(L)$?
Let me make the last part of the question more precise. Firstly, let $\mathcal X$ be the minimal regular model of $X$ over $O_K$. When I say that $x$ does not intersect $X(L)$ I mean that the intersection product $(x,y)_{\mathcal X}$ on $\mathcal X$ equals zero for all $y\in X(L)-\{x\}$. (Here $x$ and $y$ also denote their Zariski closures in $\mathcal X$.)
I think it could happen that some $K$-rational point does not intersect any other $K$-rational point. Take for example a curve with $X(K) = \{pt\}$. For some reason I do think that there should always be some (other) $L$-rational point which intersects this $K$-rational point.
The answer is no (for any genus). Fix a closed point $s\in \mathrm{Spec}(O_K)$ such that $\mathcal X_s$ is smooth. Let $x_s$ be the intersection point of $\overline{\{ x\}}$ with the fiber $\mathcal X_s$. Look (Zariski) locally at $\mathcal X$ around $x_s$. Lift a generator of the maximal ideal of $O_{\mathcal{X}_s, x_s}$ to $O_{\mathcal X, x_s}$, then you get a horizontal curve in $\mathcal X\otimes O_{S,s}$ passing through $x_s$. Taking the Zariski closure of the horizontal curve in $\mathcal X$ then gives you a horizontal curve in $\mathcal X$ passing through $x_s$. Its intersection with $\overline{\{ x\}}$ is positive, and its generic fiber is a point of $X(\bar{K})$ hence of $X(L)$ for some finite extension $L/K$.
Finally, there are infinitely many local liftings of a given generator (just perturb a lifting by adding a multiple of the uniformizing element of $O_{S,s}$), so at least one of them will gives a horizontal curve different from $\overline{\{ x\}}$.