Why does a pointed surface minus a countable set of points contain a curve? Let $S$ be a surface over $\mathbb{C}$ and let $s_1,\ldots, s_n$ be closed points of $S$.  We consider this data as fixed.
It is not hard to see that there is a curve passing through $s_1,\ldots,s_n$. I am wondering why the following is true:


For any (countable) sequence of points $x_1, x_ 2, \ldots$ in $S(\mathbb{C})\setminus \{s_1,\ldots,s_n\}$ there is an algebraic curve $C\subset S$ with  $\{s_1,\ldots,s_n\}\subset C$ and $C\cap \{x_1,x_2,\ldots \} = \emptyset$ (i.e., $C\subset S-\{x_1,\ldots\}$).


I asked a similar question on MO a while ago (see https://mathoverflow.net/questions/322084/does-there-exist-a-curve-which-avoids-a-given-countable-union-of-small-subsets) but I'm now wondering about a possible generalization of the answer given there. 
 A: The answer is based on the following fact: Let $X$ be a complex variety of positive dimension and $(Y_i)_{i \in \mathbb{N}}$ are  subvarieties of $X$ of strictly less dimension. Then $X \setminus \left( \bigcup_{i} Y_i \right)$ is non-empty (and even uncountable!).
Now assume that you have found a family $\mathcal{Z}$ of curves on $S$, such that 


*

*$\dim \mathcal{Z} \ge 1$;

*Each curve from $\mathcal{Z}$ passes through $s_1, \ldots, s_n$;

*For any $y  \in S$ which is distinct from $s_1, \ldots, s_n$,  the subvariety $\mathcal{Z}_y \subset \mathcal{Z}$ of curves passing through $y$ has strictly smaller dimension.


Then we can apply our fact to $X : = \mathcal{Z}$ and $Y_i := \mathcal{Z}_{x_i}$. Any curve corresponding to a point from $\mathcal{Z} \setminus \left ( \bigcup_i \mathcal{Z}_{x_i} \right)$ will satisfy your requirements.
In order to find this family let us take a divisor $D$ on $S$, such that the space of global sections $H^0(S, D)$ has dimension $N+1 > n+2 $ (the existence of such divisor is guaranteed by the assumption that your surface is algebraic! This statement is certainly not true for non-algebraic complex surfaces).
Observe, that then the linear map $|D|$ maps $S$ to a projective space of dimension greater than $n+1$, so for any $n+1$ point on $S$ there exists a curve $C \in |D|$ which passes through them.  
The curves in linear system $|D|$ are thus parametrised by points of the projective space $\mathbb{P}^N$. Let $\mathcal{I} \subset \mathbb{P}^N \times S^{n+1}$ be the space of collections $\{ [C] \in |D|, y_1, \ldots, y_{n+1}\in S | y_j \in C \}$. Observe that the dimension of $\mathcal{I}$  is $N+ n+1$, since it is fibered other $\mathbb{P}^N$ with the fibre being the $n+1$-th power of a curve. At the same time $\mathcal{I}$ projects to $S^{n+1}$ surjectively --- this equivalent to the observation from the previous paragraph.
Put $$\mathcal{Z}:= \mathcal{I} \bigcap \left ( \mathbb{P}^N  \times \{s_1\}\times \ldots \times \{s_n\} \times S \right )\subset \mathbb{P}^N \times S^{n+1}$$
By construction all curves in $\mathcal{Z}$ pass through $s_1, \ldots, s_n$.The dimension of $\mathcal{Z}$ is non less than $N+n+1 - 2n = N+1-n$. Now, if $\pi \colon \mathcal{Z} \to S$ is the natural projection (induced by the projection $\mathbb{P}^N  \times \{s_1\} \times \ldots \{s_n\} \times S \to S$), then $\mathcal{Z}_y = \pi^{-1}(y)$. Its dimension is $N+1-n-2=N-n-1$. By the discussion above the family $\mathcal{Z}$ containes the curve we're looking for.
