Let $R$ be the quotient of a polynomial ring $k[x_1,x_2,\dots]$ in infinitely many variables over a field by the ideal generated by all products $x_ix_j$ for $i\neq j$. Note that if $P\subset R$ is a prime ideal, then there can be at most one $i$ such that $x_i\not\in P$. It follows that if $P_i$ is the ideal generated by all the $x_j$ for $j\neq i$, every minimal prime of $R$ is of the form $P_i$ (each $P_i$ is prime since $R/P_i\cong k[x_i]$). All of these minimal primes are contained in the maximal ideal $M$ generated by all the $x_i$. So $\operatorname{Spec} R$ has a point $M$ which is in all of the infinitely many irreducible components of $\operatorname{Spec} R$. (Geometrically, you should think of $\operatorname{Spec} R$ as infinitely many lines meeting at the point $M$. Indeed, there is a natural bijection between $\operatorname{Spec} R$ and infinitely many copies of $\mathbb{A}^1_k$ with the origin in all of them identified.)

On the other hand, it is impossible for the local ring at an intersection point of infinitely many irreducible components to be a domain. To see this, we may assume our scheme is affine; say it is $\operatorname{Spec} R$ for some ring $R$, and the intersection point is some prime $P\subset R$. The irreducible components of $\operatorname{Spec} R$ correspond to the minimal prime ideals of $R$, so there are infinitely many minimal primes of $R$ contained in $P$. But there is an inclusion-preserving bijection between the prime ideals of the localization $R_P$ and the prime ideals of $R$ that are contained in $P$. So every minimal prime of $R$ corresponds to a minimal prime of $R_P$, and $R_P$ must therefore have infinitely many minimal primes. In particular, it cannot be a domain.