Non-reduced locus of a locally Noetherian scheme is the closure of non-reduced associated points Affine case (already proved by myself): The locus on $\operatorname{Spec} A$ of points where the stalks are nonreduced is the closure of those associated points of $\operatorname{Spec} A$ where the stalks are nonreduced.
I have trouble generalizing this to a locally Noetherian scheme $X$.
Let's first cover $X$ with open affines $U_i=\operatorname{Spec} A_i$ (this cover is not necessarily finite!). The associate points of $X$ are also associated in whatever $U_i$ containing them. The locus $L=\cup_iL_i$, where $L_i=U_i\cap L$. By the affine case, we know each $L_i$ is the closure those associated points of $U_i$ where the stalks are nonreduced. However, I don't know how to show that $L$ is the closure of associated points (it appears to be unions of closure instead of closure of union because of the lack of finiteness).
 A: It is a pure topological fact that the closure of a set can be computed locally:
if $X$ is a topological space, and $U_i$ is an open cover (not necessarily finite), and $S$ is any subset of $X$, then
$$
\bigcup_i((\overline{S\cap U_i})\cap U_i)=\overline{S}.
$$
Obviously the right hand side contains the left hand side. On the other hand, if $x\in\bar{S}$, then assume $x\in U_i$, and $x\in \overline{S\cap U_i}$, so $x$ is contained in the left hand side.
Now apply this to the question above. Assume $N$ is the set of nonreduced associated points, and $L$ is the non-reduced locus in $X$, then by the affine cases, we know $L\cap U_i=L_i$ equals $(\overline{N\cap U_i})\cap U_i$. Then we can see
$$
\bigcup_i L_i=\bigcup_i (\overline{N\cap U_i})\cap U_i=\overline{N}
$$ that is, the non-reduced locus is the closure of the set of nonreduced associated points.
Moreover, by the local finiteness of the set of associated points, we actually know any $x\in X$ is contained in the closure of the set of nonreduced associated points, if and only if it is contained in the closure of some non-reduced associated point $p\in X$.
(It seems my last answer has been deleted for some reason I do not know. I would be really grateful if anyone could tell me the reason. I have tried to edit that answer to make it more clarified, and then I posted it again. If this deed goes against any rule of the community, please feel free to vote to delete this answer.)
