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We use the standard convention that a two-intersecting family satisfies $|A\cap B|\ge 2$ for every pair of members, including $A=B$. In particular every member has at least two elements.

An intersecting shadow lemma. If $\mathcal{F}$ is an intersecting family of $r$-sets, $r\ge 1$, thenHere is a proof, so no additional shadow theorem is being assumed. For $i<j$, an elementary set shift replaces a member $A$ containing $j$ but not $i$ by $(A\setminus \{j\})\cup \{i\}$ exactly when that replacement is not already in the family. This preserves cardinality and the intersecting family property. The only potentially troublesome pair consists of a moved member and an unmoved member $B$ containing $j$ but not $i$. But then $B$'s replacement was already present, and the two original members obtained by using that replacement would have been disjoint if the new pair were disjoint.

Alsoso an elementary set shift cannot increase the lower shadow. For completeness, a shadow set containing neither $i,j$ or both has an unchanged witness in the original lower shadow. A shadow set containing $i$ but not $j$ is either originally present or the replacement of an originally present shadow set. If a shadow set contains $j$ but not $i$, its witnessing member survived the elementary set shift; either that member contains both coordinates, or its replacement was already present. In either case both the shadow set and its replacement were originally present, so the former survives in the shifted lower shadow.

Repeated elementary set shifts terminate because the sum of the elements of all members decreases whenever the family changes. In the resulting shifted set family $\mathcal{H}$, if $a_1<\cdots <a_r$ is a member and $b_1<\cdots <b_r$ satisfies $b_i\le a_i$ for every $i$, then $\{b_1,\dots ,b_r\}$ is also a member. Every $A\in \mathcal{H}$ has a prefix $[2q-1]$ containing $q$ of its elements. Otherwise its sorted elements satisfy $a_i\ge 2i$, and the first $r$ elements outside $A$ form a disjoint member componentwise no larger than $A$, contradicting the intersecting family property.

Choose the first such prefix and reflect membership inside it:This has size $r-1$. The set $D\cup \{2q-1\}$ is componentwise no larger than $A$: before the first excess, the prefix counts of $A$ are at most half the prefix length, and reflection only moves selected positions to the left. Thus $D$ belongs to $\partial \mathcal{H}$. This ballot reflection injection is injective: in $D$, the reflected prefix ends at the first position where unselected positions outnumber selected ones, and reflecting it again recovers $A$. Consequently $|\partial \mathcal{H}|\ge |\mathcal{H}|=|\mathcal{F}|$. Since elementary set shifts did not increase the lower shadow, this proves the intersecting shadow lemma.

The nonuniform two-intersecting family bound. Write ${\mathcal{A}}_i=\mathcal{A}\cap [n{]}^{(i)}$. No member of $\partial {\mathcal{A}}_i$ can be the complement of a member of ${\mathcal{A}}_{n-i+1}$: such an inclusion would make the intersection of the two original members have at most one element. The intersecting shadow lemma therefore givesFor $n=2h$, sum this over $1\le i\le h$. These pairs cover all nonempty levels, soEquality is attained by all subsets of size at least $h+1$: any two intersect in at least two elements.

For $n=2h+1$, the same pairs cover every level except $h+1$. Complements of the members of ${\mathcal{A}}_{h+1}$ form an intersecting family of $h$-sets, sinceFor $h\ge 1$, the permitted Erdős-Ko-Rado theorem bounds this middle level by $\left(\genfrac{}{}{0ex}{}{2h}{h-1}\right)=\left(\genfrac{}{}{0ex}{}{2h}{h+1}\right)$. HenceEquality is attained by all $(h+1)$-sets inside $[2h]$, together with all subsets of $[2h+1]$ of size at least $h+2$. Two members of the first group intersect in at least two elements; a member from each group intersects in at least two; two members of the second group intersect in at least three. For $n=1$ the only two-intersecting family is empty, which also gives equality. The empty-ground-set case is likewise immediate.

The pair-cover bound for a two-intersecting uniform family. If $\mathcal{A}$ is nonempty, fix $A_0\in \mathcal{A}$, with $|A_0|=r$. Every member contains one of the $\left(\genfrac{}{}{0ex}{}{r}{2}\right)$ pairs inside $A_0$. Each fixed pair belongs to exactly $\left(\genfrac{}{}{0ex}{}{n-2}{r-2}\right)$ $r$-sets, so counting their union givesThe empty family is immediate; a nonempty two-intersecting family requires $r\ge 2$.

Witness-pair concentration for an intersecting uniform family. Let $\mathcal{A}$ now be an intersecting family of $r$-sets. If it is two-intersecting, the preceding bound, together with $\left(\genfrac{}{}{0ex}{}{r}{2}\right)\le (r-1{)}^2$ for $r\ge 2$, shows that any chosen $x\in [n]$ has the required property. Otherwise there are members $A,B$ with $A\cap B=\{x\}$. Every member $C$ not containing $x$ must contain some $a\in A\setminus \{x\}$ and some $b\in B\setminus \{x\}$. These two sets of possible witnesses are disjoint. There are $(r-1{)}^2$ witness pairs, and each is contained in $\left(\genfrac{}{}{0ex}{}{n-2}{r-2}\right)$ $r$-sets. ThusFor $r=1$, a nonempty intersecting family consists of one singleton; choose its element and the number avoiding it is zero. This is the natural separate interpretation of the degenerate case, where the printed binomial coefficient has lower index $-1$. For an empty family any $x\in [n]$ works, provided $n\ge 1$ as usual.