All schemes are Noetherian. Consider a scheme $X$ and an open subscheme $U\subseteq X$. Let $\partial U$ denote the boundary as a topological subspace. I want to justify the following statement:
For any closed point $x\in\partial U$, there exists a 1-dimensional integral subscheme $C\hookrightarrow X$ such that $C\cap U\ne\emptyset$ and $x\in C$.
If the above is true I wonder whether we can assume further that
$C\cap\partial U=\{x\}$.
Thanks.
This is only a partial answer.
If you are OK with restricting to the case of varieties, then any two points of a variety $X$ are connected by a curve, so you can just connect $x$ to a point of $U$.
For the second question the answer is no: just take $X=\mathbb{P}^1$, $U=\mathbb{P}^1\setminus\{p,q\}$ for some distinct points $p$ and $q$. The only choice for $C$ is $X$. I do not know what happens in higher dimensions, i.e. for schemes with no 1-dimensional irreducible components.