This theorem seems "obvious" to me, but I want to check my logic since I am un-familiar with un-countably infinite sets, and I know these can give rise to non-intuitive results. Any comments welcome.
Definitions
The cardinality of a set $A$ will be denoted $|A|$. If $|A|=|\mathbb{N}|$, $A$ will be called countable or countably infinite and this will be denoted $|A|=\aleph_{0}$. If $A$ is infinite and $|A|\ne\aleph_{0}$, $A$ will be called uncountable or uncountably infinite.
Claim
If $A$ is un-countable then for any $a\in A$, $A$ contains an uncountable number of disjoint co-countable sets that contain $a$. More formally;
For any $a\in A$ where $|A|\not=\aleph_{0}$ there exist subsets $B_{\beta}\subset A$, $\beta\in I_{\beta}$ for an index set $I_{\beta}$, $|I_{\beta}|\not=\aleph_{0}$, such that for all $\beta\in I_{\beta}$ : $a\in B_{\beta}$ and $|A-B_{\beta}|=\aleph_{0}$. Furthermore for any $\beta\not=\beta'$, $B_{\beta}\cap B_{\beta'}=\emptyset$
Proof
Without loss of generality let $I_{\beta}=\mathbb{R}$ and choose $\beta\in I_{\beta}$. Choose $a_{\beta,1}\in A$, $a_{\beta,1}\not=a$. Since $|A-\{a_{\beta,1}\}|\not=\aleph_{0}$ there exists $a_{\beta,2}\in\{A-\{a_{\beta,1}\}\}$, $a_{\beta,2}\not=a$. Since $|A-\{a_{\beta,1},a_{\beta,2}\}|\not=\aleph_{0}$ there exists $a_{\beta,3}\in\{A-\{a_{\beta,1},a_{\beta,2}\}\}$, $a_{\beta,3}\not=a$. Continuing in this way $n_{\beta}$ times, $n_{\beta}\in\mathbb{N}$, we can construct a set $B_{\beta}=A-\{a_{\beta,1},a_{\beta,2},...,a_{\beta,n_{\beta}}\}$ that satisfies $a\in B_{\beta}\subset A$ and $|A-B_{\beta}|=|\{a_{\beta,1},a_{\beta,2},...,a_{\beta,n_{\beta}}\}|=\aleph_{0}$ so that $B_{\beta}$ is a co-countable subset of $A$ containing $a$.
Choose another $\beta'\in I_{\beta}$, $\beta\not=\beta'$. Since $|B_{\beta}|\not=\aleph_{0}$ there exists $a_{\beta',1}\in B_{\beta}$, $a_{\beta',1}\not=a$. Since $|B_{\beta}-\{a_{\beta',1}\}|\not=\aleph_{0}$ there exists $a_{\beta',2}\in B_{\beta}-\{a_{\beta',1}\}$, $a_{\beta',2}\not=a$. Since $|B_{\beta}-\{a_{\beta',1},a_{\beta',2}\}|\not=\aleph_{0}$ there exists $a_{\beta',3}\in B_{\beta}-\{a_{\beta',1},a_{\beta',2}\}$, $a_{\beta',3}\not=a$. Continuing in this way $n_{\beta'}$ times, $n_{\beta'}\in\mathbb{N}$, we can construct a set $B_{\beta'}=B_{\beta}-\{a_{\beta',1},a_{\beta',2},...,a_{\beta',n_{\beta'}}\}$ that satisfies $a\in B_{\beta'}\subset A$ and $|A-B_{\beta'}|=|\{a_{\beta,1},a_{\beta,2},...,a_{\beta,n_{\beta}}\}\cup\{a_{\beta,1},a_{\beta,2},...,a_{\beta',n_{\beta'}}\}|=\aleph_{0}+\aleph_{0}=\aleph_{0}$ so that $B_{\beta'}$ is a co-countable subset of $A$ containing $a$.
Since $B_{\beta}\cap B_{\beta'}=\emptyset$ and since we can choose un-countably many $\beta\in I_{\beta}$ the result is proved $\hspace{5pt}\blacksquare$