The is a model structure on simplicial sets whose fibrant objects are "quasicategories", the simplicial sets in which every inner horn has a filler. If the 0-simplices are thought of as objects of a category and the 1-simplices as morphisms, then the horn filler condition gives a sense in which a quasicategory admits composites of all arrows, with composition associative up to a homotopy, which is itself well defined up to a higher homotopy, which is...and so on. The latter is the primeval concept of an $(\infty,1)$-category, and quasicategories are the most used model.
In fact the first model category for $(\infty,1)$-categories that was introduced was that of Bergner, which is on the category of simplicially enriched categories. This models an $(\infty,1)$-category as a "category" with morphisms of all dimensions, all admitting strictly associative compositions. It is very far from obvious that everything we'd like to call an $(\infty,1)$-category is equivalent to something of this form, but in fact this is the case: there is a Quillen equivalence between the model categories of simplicial categories and of quasicategories, showing that every $(\infty,1)$-category (viewed as a quasicategory) can be viewed as a simplicially enriched category.
The sense in which a quasicategory $Q$ is "equivalent" to some simplically enriched category $\mathcal C$ is that the homotopy categories $\mathrm{Ho}(Q)$ and $\mathrm{Ho}(\mathcal C)$ are equivalent, but also, roughly speaking, that
$Q$ and $\mathcal C$ have the same mapping spaces. One can't formalize this directly by asking that there be a map inducing such equivalences $Q\to \mathcal C$, as they don't live in the same category, which is the reason for the introduction of the Quillen equivalence between the model structures.
Maybe it would be easier to see that $\langle \pi_1,\pi_2\rangle:A\times_BA\rightarrow A\times A$ is an internal equivalence relation in $\mathcal{C}$.
One way of explaining this is the following. First note that this is obvious if $\mathcal{C} = \textbf{Set}$. Indeed, $A\times_BA$ consists of pairs $(a_1,a_2)\in A\times A$ such that $p(a_1) = p(a_2)$ and this clearly gives rise to an equivalence relation on $A$. Now for general category $\mathcal{C}$ you can use essentially the same argument as for $\textbf{Set}$ if you apply Yoneda embedding.
Hence $\langle \pi_1,\pi_2\rangle:A\times_BA\rightarrow A\times A$ is an internal equivalence relation in $\mathcal{C}$. Recall that we may view every equivalence relation as a rather simple groupoid and thus as a category. Therefore, $\langle \pi_1,\pi_2\rangle:A\times_BA\rightarrow A\times A$ admits a structure of an internal category.
Best Answer
I would add to the current answer that you seem to be confusing 3 different things, which each are a category associated to $\mathbb{N}$:
the discrete category associated to $\mathbb{N}$: for any set $X$, you can take a category whose objects are the elements of $X$, and the only morphisms are the identity of each object;
the category associated to $\mathbb{N}$ as an ordered set: for any (partially) ordered set $(X,\leqslant)$, you can define a category whose objects are the elements of $X$, and there is a unique morphism $x\to y$ iff $x\leqslant y$;
the category associated to $\mathbb{N}$ as a topological space: for any topological space $X$, you can define the category associated (like in the previous point) to the ordered set $\mathcal{O}(X)$ of open subsets of $X$, with the inclusion. That is, the objects of this category are the open subsets of $X$, and there is a morphism $U\to V$ iff $U\subset V$.
When applied to $\mathbb{N}$, this gives three different categories, although the first one is a subcategory of the second one (they have the same objects, but the second one has more morphisms), which can also be seen as a subcategory of the third one by sending $n\mapsto \{0,\dots,n\}$.
Note that since $\mathbb{N}$ is a discrete topological space for its canonical topology, the objects of the third category above actually are all subsets of $\mathbb{N}$. On the other hand, it is not at all a discrete category.
Fun fact: $\mathbb{N}$ can also be called a discrete ordered set. This does not imply that any of the second or third categories above is discrete, but it does imply that $\mathbb{N}$ is discrete as a topological space. Words are funny.