Let $X$ a complex curve and $x\in X$ a point.
We consider the space of effective divisors $D$ with fixed degree $d$, whic we know is isomorphic to $X^{d}/S_{d}$ where $S_{d}$ is the symmetric group.
Now, we consider the subspace of divisors $D$ with fixed degree $d$ such that:
$\sum\limits_{x untransversal}m_{x}(D)\leq N.$
where the untransversal points means that $m_{x}(D)\geq 2$.
Is this space open in $X^{d}/S_{d}$?
First of all, I assume you're asking about all effective divisors of degree $d$ that satisfy that inequality; if not, then it doesn't make sense to talk about them in $X^d/S_d$.
Assuming this, then you're asking about all the $(x_1,\ldots,x_d)\in X^d/S_d$ such that some "coordinates" are the same, and such that the amount of coordinates that are the same do not add up to more than $N$. This space is open in $X^d/S_d$, for the following reason:
We have the morphism $p_{ij}:X^d/S_d\to X^2/S_2$, where $x_1+\cdots+x_d\mapsto x_i+x_j$. Let $\Delta$ denote the "diagonal" in $X^2/S_2$; that is $\Delta=\{2x:x\in X\}$. This is closed in $X^2/S_2$.
Assume first that we want the set of divisors such that the sum of untransversal points is equal to 2. This is then the closed set $\bigcup_{i,j}p_{ij}^{-1}(\Delta)$.
For similar reasons (which involve more complicated formulas using unions and intersections of the inverse images of these diagonals that I don't feel like writing right now), the set of divisors such that the sum of untransversal points is equal to a given $m\in\mathbb{N}$ (following a user's comment below, let's name these $\mbox{Unt}_m$) is also closed. The set of divisors such that the sum of untransversal points is less than or equal to $N$ is the complement of the union $\bigcup_{N\leq m\leq d}\mbox{Unt}_m$, and so is open.
