I think this is a consequence of (variants of) Hilbert's irreducibility theorem. Let me explain why. Suppose that $C$ is geometrically integral over a number field $K$. Let $K'/K$ be a normal finite extension. We will show that infinitely many points of $C$ are defined over a field disjoint from $K'$.

Since both the hypotheses and the conclusion are birational invariants, we may suppose that $C$ is a closed subset of $\mathbb{A}_K^2$ (take an affine open of $C$, embed it in $\mathbb{A}^N$ for some $N$ and take a generic projection to $\mathbb{A}^2$). Choose a generic projection $p:C\to\mathbb{A}^1_K$. The curve $C$ is described by an equation $F(t,x)=0$, where $t$ is the coordinate of $\mathbb{A}^1_K$, and $F$ is an irreducible polynomial.

Now, since $C$ is geometrically integral, $F_{K'}$ is still irreducible. By [Serre, Topics in Galois theory, Proposition 3.3.1], $\lambda'\mapsto F_{K'}(\lambda',x)$ is irreducible for every $\lambda'\in K'$ outside of a thin set. Hence, by [Serre, Topics in Galois theory, Proposition 3.2.1], $\lambda\mapsto F_{K'}(\lambda,x)$ is irreducible for every $\lambda\in K$ outside of a thin set. Since $K$ is Hilbertian, this holds for infinitely many $\lambda\in K$. For such a $\lambda$, $p_{K'}^{-1}(\lambda)\subset C_{K'}$ consists of a unique (reduced) point  $p'\in C_{K'}$. Let us denote by $p$ its image in $C$, and let $L$ and $L'$ be the residual fields of $p$ and $ p'$. By construction, $L'=L\otimes_K K'$which implies that $L$ is disjoint from $K'$.

I think this is a consequence of (variants of) Hilbert's irreducibility theorem. Let me explain why. Suppose that $C$ is a geometrically integral curve defined over a number field $K$. Let $K'/K$ be a normal finite extension. We will show that infinitely many points of $C$ are defined over a field disjoint from $K'$.

Since both the hypotheses and the conclusion are birational invariants, we may suppose that $C$ is a closed subset of $\mathbb{A}_K^2$ (take an affine open of $C$, embed it in $\mathbb{A}^N$ for some $N$ and take a generic projection to $\mathbb{A}^2$). Choose a generic projection $p:C\to\mathbb{A}^1_K$. The curve $C$ is described by an equation $F(t,x)=0$, where $t$ is the coordinate of $\mathbb{A}^1_K$, and $F$ is an irreducible polynomial.

Now, since $C$ is geometrically integral, $F_{K'}$ is still irreducible. By [Serre, Topics in Galois theory, Proposition 3.3.1], $x\mapsto F_{K'}(\lambda',x)$ is irreducible for every $\lambda'\in K'$ outside of a thin set. Hence, by [Serre, Topics in Galois theory, Proposition 3.2.1], $x\mapsto F_{K'}(\lambda,x)$ is irreducible for every $\lambda\in K$ outside of a thin set. Since $K$ is Hilbertian, this holds for infinitely many $\lambda\in K$.

Let us fix such a $\lambda$. We denote by $q$ and $q'$, the points of $\mathbb{A}^1_K$ and $\mathbb{A}^1_{K'}$ with coordinate $\lambda$. By choice of $\lambda$, $x\mapsto F_{K'}(\lambda,x)$ hence also $x\mapsto F(\lambda,x)$ are irreducible polynomials. Hence $p_{K'}^{-1}(q')\subset C_{K'}$ (resp. $p^{-1}(q)\subset C$) consists of a unique (reduced) point  $p'\in C_{K'}$ (resp. $p\in C$). Let $L$ and $L'$ be the residual fields of $p$ and $p'$. By construction, $p'=p\times_{q} q'$ so that $L'=L\otimes_K K'$. This implies that $L$ is disjoint from $K'$.