Let $\Omega = (\mathbb{N}_{\infty} \xrightarrow{p} \mathbb{N}_{\infty} \xrightarrow{p} \dotsc)$ be as described.
Let $S \subseteq X$ be a subobject, thus we have a bunch of compatible injections $S_i \to X_i$. Compatibility means that the diagrams
$$\begin{array}{c} X_i & \rightarrow & X_{i+1} \\ \downarrow && \downarrow \\ S_i & \rightarrow & S_{i+1} \end{array}$$
commute.
Define $\phi : X \to \Omega$ as follows: If $i \in \mathbb{N}$, we want to define $\phi_i : X_i \to \Omega_i = \mathbb{N}_{\infty}$. Well, if $x \in X_i$, then there are three cases:
$x \in S_i$ (by which I mean that $x$ lies in the image of $S_i \to X_i$). Then $\phi_i(x):=0$.
More generally, assume that the image of $x$ in $X_{i+n}$ lies in $S_{i+n}$ for some $n \geq 0$. Choose $n$ minimal. Then $\phi_i(x) := n$.
Otherwise, we define $\phi_i(x) := \infty$.
By the very construction, the diagram
$$\begin{array}{c} X_i & \rightarrow & X_{i+1} \\ \phi_i \downarrow ~~~~ && ~~~~ \downarrow \phi_{i+1} \\ \mathbb{N}_\infty & \xrightarrow{p} & \mathbb{N}_\infty \end{array}$$
commutes, i.e. $\phi : X \to \Omega$ is a morphism. One can also check that we have a pullback diagram, as desired.
I think you've seen why it ought to be the intersection, but for the sake of completeness I'll be especially painstaking about it.
Let's just use $i:E\rightarrowtail X$ for the subobject classified by $F\circ\langle\phi,\psi\rangle$, and $i_A,i_B$ be the inclusion morphisms of $A,B$, respectively, into $X$.
By the lefthand commutative square in the above diagram, we have that $\phi\circ i=\top\circ !_E$ (so $i$ factors through $i_A$) and $\psi\circ i=\top\circ !_E$ (so $i$ factors through $i_B$). It follows from this (along with the fact that $i$ is a monomorphism) that $E$ is a subobject of both $A$ and $B$, and this is our first clue that we're looking at what ought to be an intersection. Let's denote by $j_A,j_B$ the respective inclusions of $E$ into $A,B$; note, in particular, that $i_Aj_A=i=i_Bj_B$.
Now let $f:Q\to A$ and $g:Q\to B$ be morphisms such that $i_Af=i_Bg$; we wish to show that there is a unique map $h:Q\to E$ with $j_A\circ h=f$ and $j_B\circ h=g$; that is, we want to show that $E$ is the pullback of $i_A$ and $i_B$.
Denote by $\alpha$ the morphism $i_Af=i_Bg:Q\to X$. Then $$\phi\circ\alpha=\phi\circ i_A\circ f=\top\circ !_A\circ f=\top\circ!_Q$$ and $$\psi\circ\alpha=\psi\circ i_B\circ g=\top\circ !_B\circ g=\top\circ!_Q,$$ meaning that $\langle\phi,\psi\rangle\circ\alpha=\langle\top,\top\rangle\circ !_Q$. Because that left hand square is a pullback, there is a unique morphism $h:Q\to E$ with $i\circ h=\alpha$. This equality gives us both $$i_A\circ j_A\circ h=i\circ h=\alpha=i_A\circ f$$ from which we have $j_Ah=f$ because $i_A$ is a monomorphism, and $$i_B\circ j_B\circ h=i\circ h=\alpha=i_B\circ g$$ from which we have $j_B\circ h=g$.
Thus we've shown that $E$ satisfies the universal property of the intersection of $A$ and $B$, using only that the squares in the pictured diagram are pullbacks.
Best Answer
Thinking about maps to $\Omega$ as subobjects, this is just saying that if you have a subobject $C\subseteq A$, then there is a subobject $D\subseteq B$ such that $C=A\cap D$ as subobjects of $B$. How do you find such a $D$? Well, you can just take $D=C$!
To make this precise, let $i:C\to A$ be the subobject classified by $f$. Then $gi:C\to B$ is monic since $i$ and $g$ are, so it is classified by a map $h:B\to\Omega$. Now in the diagram $$\require{AMScd} \begin{CD} C @>{1}>> C @>{}>> 1\\ @V{i}VV @VV{gi}V @VV{t}V\\ A @>{g}>> B @>{h}>> \Omega \end{CD} $$ the right square is a pullback by definition of $h$ and the left square is a pullback since $g$ is monic. This implies the outer rectangle is a pullback, i.e. that $hg$ classifies the subobject $i:C\to A$, so $hg=f$.