[proofplan]
We first reduce to the case in which every member of the family has the same finite cardinality. The proof then proceeds by induction on that common cardinality. In the induction step, either one point occurs in uncountably many members, in which case we delete that point and apply the induction hypothesis, or no point occurs uncountably often, in which case we recursively build an uncountable pairwise disjoint subfamily. The first case produces a nonempty root by adding the common point back; the second case produces a delta-system with empty root.
[/proofplan]
[step:Thin the family to sets of one fixed finite size]
Let $\mathbb{N}$ denote the set of positive integers. For each $n \in \mathbb{N} \cup \{0\}$, define
\begin{align*}
\mathcal{A}_n := \{A \in \mathcal{A} : |A| = n\}.
\end{align*}
Then
\begin{align*}
\mathcal{A} = \bigcup_{n=0}^{\infty} \mathcal{A}_n.
\end{align*}
If every $\mathcal{A}_n$ were countable, then $\mathcal{A}$ would be a [countable union of countable sets](/theorems/755), hence countable, contradicting the hypothesis that $\mathcal{A}$ is uncountable. Therefore there exists $n \in \mathbb{N} \cup \{0\}$ such that $\mathcal{A}_n$ is uncountable.
It is enough to prove the theorem under the additional hypothesis that every member of $\mathcal{A}$ has cardinality $n$, because an uncountable delta-subfamily of $\mathcal{A}_n$ is also an uncountable delta-subfamily of $\mathcal{A}$.
[/step]
[step:Prove the fixed-cardinality statement by induction]
For $n \in \mathbb{N} \cup \{0\}$, let $P(n)$ be the assertion that every uncountable family of sets of cardinality $n$ contains an uncountable delta-system with finite root.
The assertion $P(0)$ holds: if every member of $\mathcal{A}$ has cardinality $0$, then every member is $\varnothing$, and any uncountable subfamily is a delta-system with root $\varnothing$.
Assume $n \in \mathbb{N} \cup \{0\}$ and assume $P(n)$. Let $\mathcal{A}$ be an uncountable family of sets of cardinality $n+1$. We prove that $\mathcal{A}$ contains an uncountable delta-system.
[/step]
[step:If one element appears uncountably often, delete it and use the induction hypothesis]
Suppose there exists a set element $x$ such that
\begin{align*}
\mathcal{A}_x := \{A \in \mathcal{A} : x \in A\}
\end{align*}
is uncountable. Define
\begin{align*}
\mathcal{C} := \{A \setminus \{x\} : A \in \mathcal{A}_x\}.
\end{align*}
The map $\Phi: \mathcal{A}_x \to \mathcal{C}$ defined by $\Phi(A) := A \setminus \{x\}$ is injective, because $A = \Phi(A) \cup \{x\}$ for every $A \in \mathcal{A}_x$. Hence $\mathcal{C}$ is uncountable. Each member of $\mathcal{C}$ has cardinality $n$.
By the induction hypothesis $P(n)$, there exist an uncountable subfamily $\mathcal{D} \subset \mathcal{C}$ and a finite set $r$ such that for all distinct $D,E \in \mathcal{D}$,
\begin{align*}
D \cap E = r.
\end{align*}
Define
\begin{align*}
\mathcal{B} := \{D \cup \{x\} : D \in \mathcal{D}\}.
\end{align*}
Then $\mathcal{B}$ is an uncountable subfamily of $\mathcal{A}_x \subset \mathcal{A}$. Since no member of $\mathcal{C}$ contains $x$, we have $x \notin r$. Therefore, for distinct $D,E \in \mathcal{D}$,
\begin{align*}
(D \cup \{x\}) \cap (E \cup \{x\}) = (D \cap E) \cup \{x\} = r \cup \{x\}.
\end{align*}
Thus $\mathcal{B}$ is an uncountable delta-system with finite root $r \cup \{x\}$.
[guided]
Assume that some element $x$ belongs to uncountably many members of the original family. We isolate exactly those members by defining
\begin{align*}
\mathcal{A}_x := \{A \in \mathcal{A} : x \in A\}.
\end{align*}
The reason to pass to $\mathcal{A}_x$ is that $x$ is now a guaranteed common point of every set under consideration. To use the induction hypothesis, we remove this common point. Define
\begin{align*}
\mathcal{C} := \{A \setminus \{x\} : A \in \mathcal{A}_x\}.
\end{align*}
Every member of $\mathcal{C}$ has cardinality $n$, because every member of $\mathcal{A}_x$ has cardinality $n+1$ and contains $x$.
We must also check that $\mathcal{C}$ is uncountable. The deletion map $\Phi: \mathcal{A}_x \to \mathcal{C}$ defined by $\Phi(A) := A \setminus \{x\}$ is injective: if $\Phi(A) = \Phi(B)$, then
\begin{align*}
A = \Phi(A) \cup \{x\} = \Phi(B) \cup \{x\} = B.
\end{align*}
Since $\mathcal{A}_x$ is uncountable and injects into $\mathcal{C}$ by a surjective definition of $\mathcal{C}$, the family $\mathcal{C}$ is uncountable.
Now the induction hypothesis applies to $\mathcal{C}$. Hence there are an uncountable subfamily $\mathcal{D} \subset \mathcal{C}$ and a finite set $r$ such that
\begin{align*}
D \cap E = r
\end{align*}
for all distinct $D,E \in \mathcal{D}$. Adding $x$ back to every set gives
\begin{align*}
\mathcal{B} := \{D \cup \{x\} : D \in \mathcal{D}\}.
\end{align*}
This family is uncountable and lies inside $\mathcal{A}$. Also, $x$ is not in any member of $\mathcal{C}$, so $x \notin r$. Therefore, whenever $D,E \in \mathcal{D}$ are distinct,
\begin{align*}
(D \cup \{x\}) \cap (E \cup \{x\}) = (D \cap E) \cup \{x\} = r \cup \{x\}.
\end{align*}
Thus $\mathcal{B}$ is a delta-system with finite root $r \cup \{x\}$.
[/guided]
[/step]
[step:If no element appears uncountably often, recursively build disjoint sets]
It remains to handle the case in which, for every set element $x$,
\begin{align*}
\{A \in \mathcal{A} : x \in A\}
\end{align*}
is countable.
Let $\omega_1$ denote the first uncountable ordinal, so every ordinal $\alpha < \omega_1$ is countable and the index set $\{\alpha : \alpha < \omega_1\}$ is uncountable. We construct by transfinite recursion a family $(A_\alpha)_{\alpha < \omega_1}$ of members of $\mathcal{A}$ such that
\begin{align*}
A_\alpha \cap A_\beta = \varnothing
\end{align*}
whenever $\alpha < \beta < \omega_1$.
Assume $(A_\beta)_{\beta < \alpha}$ has been chosen for some countable ordinal $\alpha < \omega_1$. Define
\begin{align*}
U_\alpha := \bigcup_{\beta < \alpha} A_\beta.
\end{align*}
Since $\alpha$ is countable and each $A_\beta$ is finite, the set $U_\alpha$ is countable. Define the excluded subfamily
\begin{align*}
\mathcal{E}_\alpha := \{A \in \mathcal{A} : A \cap U_\alpha \neq \varnothing\}.
\end{align*}
For every $x \in U_\alpha$, the family $\{A \in \mathcal{A} : x \in A\}$ is countable by the present case assumption. Hence
\begin{align*}
\mathcal{E}_\alpha \subset \bigcup_{x \in U_\alpha} \{A \in \mathcal{A} : x \in A\},
\end{align*}
so $\mathcal{E}_\alpha$ is countable as a countable union of countable sets. Since $\mathcal{A}$ is uncountable, $\mathcal{A} \setminus \mathcal{E}_\alpha$ is nonempty. Choose
\begin{align*}
A_\alpha \in \mathcal{A} \setminus \mathcal{E}_\alpha.
\end{align*}
Then $A_\alpha \cap U_\alpha = \varnothing$, so $A_\alpha$ is disjoint from every earlier $A_\beta$.
The resulting family
\begin{align*}
\mathcal{B} := \{A_\alpha : \alpha < \omega_1\}
\end{align*}
is uncountable and pairwise disjoint. Therefore, for all distinct $A,B \in \mathcal{B}$,
\begin{align*}
A \cap B = \varnothing.
\end{align*}
Thus $\mathcal{B}$ is a delta-system with finite root $\varnothing$.
[/step]
[step:Conclude the induction and return to the original family]
The two cases exhaust all possibilities for an uncountable family of sets of cardinality $n+1$. Hence $P(n)$ implies $P(n+1)$. Since $P(0)$ holds, induction gives $P(n)$ for every $n \in \mathbb{N} \cup \{0\}$.
By the initial thinning step, the original uncountable family $\mathcal{A}$ contains an uncountable subfamily whose members all have one fixed finite cardinality. Applying the fixed-cardinality result to that subfamily gives an uncountable delta-system $\mathcal{B} \subset \mathcal{A}$ with finite root $r$. This proves the theorem.
[/step]
