Recall a locally Noetherian scheme $X$ has Serre's condition $S_k$ if for every $x\in X$ we have $\mathrm{depth}(\mathcal{O}_{x,X})\geq \mathrm{min}(k,\mathrm{dim}(\mathcal{O}_{x,X}))$.
Let $X$ be a variety, $Z\subset X$ be a closed subset of codimension $k+1$. My question is, suppose $X\setminus Z$ has Serre's condition $S_k$, does $X$ also have $S_k$?
I am just posting my comment as one answer. Let $k$ be a field, and let $R$ be the following (unital, commutative) $k$-algebra, $$R = k[s,t,u,x,y,z]/I, \ \ I = \langle sx,sy,sz,tx,ty,tz,ux,uy,uz \rangle.$$ As a closed subscheme of $\mathbb{A}^6_k = \text{Spec}\ k[s,t,u,x,y,z]$, the scheme $\text{Spec}(R)=\text{Zero}(I)$ is a reduced scheme that is a union of two irreducible components, each of which is a linear affine $3$-space, such that the intersection of the two irreducible components equals the singleton set of the origin, $Z=\text{Zero}(\langle s,t,u,x,y,z \rangle)$. Thus, $\text{Spec}(R)\setminus Z$ is a smooth $k$-scheme equal to the disjoint union of two copies of the open complement of the origin in affine $3$-space. In particular, this open subscheme of $\text{Spec}(R)$ is $\text{S}_2$ (in fact, Cohen-Macaulay, and local complete intersection, and Gorenstein, and regular). However, $\text{Spec}(R)$ is not $\text{S}_2$, since $\text{S}_2$-schemes are connected away from codimension $>1$.
