proof of delta system lemma

Since there are only ${\mathrm{\aleph }}_0$ possible cardinalities for any element of $S$, there must be some $n$ such that there are an uncountable number of elements of $S$ with cardinality $n$. Let $S^{*}=\{a\in S\mid |a|=n\}$ for this $n$. By induction, the lemma holds:

If $n=1$ then there each element of $S^{*}$ is distinct, and has no intersection with the others, so $X=\mathrm{\varnothing }$ and $S^{\prime }=S^{*}$.

Suppose $n>1$. If there is some $x$ which is in an uncountable number of elements of $S^{*}$ then take $S^{**}=\{a\setminus \{x\}\mid x\in a\in S^{*}\}$. Obviously this is uncountable and every element has $n-1$ elements, so by the induction hypothesis there is some $S^{\prime }\subseteq S^{**}$ of uncountable cardinality such that the intersection of any two elements is $X$. Obviously $\{a\cup \{x\}\mid a\in S^{\prime }\}$ satisfies the lemma, since the intersection of any two elements is $X\cup \{x\}$.

On the other hand, if there is no such $x$ then we can construct a sequence such that each $a_i\in S^{*}$ and for any $i\ne j$, $a_i\cap a_j=\mathrm{\varnothing }$ by induction. Take any element for $a_0$, and given , since $\alpha$ is countable, is countable. Obviously each element of $A$ is in only a countable number of elements of $S^{*}$, so there are an uncountable number of elements of $S^{*}$ which are candidates for $a_{\alpha }$. Then this sequence satisfies the lemma, since the intersection of any two elements is $\mathrm{\varnothing }$.