Prove that this is a measure I have problems with this exercise

Let $X$ be an uncountable set, and let $\Sigma \subseteq{}\mathcal{P}
 (X) $ the $\sigma$-algebra.
$    $$\boldsymbol{\beta} = \{A \subseteq{X} : A$ it is countable or
$A^c$ it is countable $\}$
Define $\mu: \boldsymbol{\beta} \to
 \overline{\mathbb{R}_{+}} $ ($\mathbb{R}_{+} = [0, + \infty]$ ) as
$\mu (A) = \left \{ \begin{matrix} 0 & \mbox{if }A\mbox{ is countable}
 \\ 1 & \mbox{if }A^c\mbox{ is countable}\end{matrix}\right. $
and prove that $\mu$ is a measure.

My attempt:
$\mu : C \rightarrow{\overline{\mathbb{R}}}$ is a measure if:
i) $\mu( \emptyset ) =0$
ii) If ${A_n}_{n \in \mathbb{N}} \subseteq{C}$ such that $ \displaystyle\bigcup_{n \in \mathbb{N}}^{}{A_n } \in C$ and the $A_n$'s are disjoint
$$\mu(\displaystyle\bigcup_{n \in \mathbb{N}}^{}{A_n }) = \displaystyle\sum_{n \in \mathbb{N}}^{}\mu(A_n)$$
i) $\mu( \emptyset ) =0$ because the cardinality of  $\emptyset = 0$
ii) Let $\{A_n\mid n \in \Bbb N\}$ a collection of sets of  $\boldsymbol{\beta}$ disjoint two by two. Note that, since two by two are disjoint, there can be at most one with a countable complement. Then there are two possibilities: either they are all countable (and therefore their union is countable) or there is exactly one with a countable complement and the others are countable (and therefore the union has countable complement).
Thanks
 A: Firstly, as $X$ is uncountable, none of it's subsets are both countable and have countable complement. This fact makes $\mu$ well defined. To prove $\mu$ is $\sigma$-additive, you can proceed using a proof by cases:
Consider a countable collection of disjoint sets $\mathcal{A}\subset\Sigma$. If all the sets in $\mathcal{A}$ are countable then you are done as a countable collection of countable sets is countable (proof), in particular: $$\mu(\bigcup_{n\in\mathbb{N}}A_n)=0=\sum_{n\in\mathbb{N}}0= \sum_{n\in\mathbb{N}}\mu(A_{n}).$$ Suppose then that there exists an $m\in\mathbb{N}$ such that $X\setminus A_m$ is countable. Thus we have that all the sets in $\{A_n\}_{n\ne m}$ are countable as $$\{A_n\}_{n\ne m}\subset\mathcal{P}(X\setminus A_m)$$
since $\mathcal{A}$ is a disjoint collection. Therefore we have $$\sum_{n\in\mathbb{N}}\mu(A_n)=\mu(A_m)+\sum_{n\ne m}\mu(A_n)=1+\sum_{n\ne m}0=1.$$ Also, $X\setminus\bigcup\mathcal{A}$ is countable as $$X\setminus \big[(\bigcup_{n\ne m} A_n) \cup A_m \big]\subset X\setminus A_m$$ giving $$\mu(\bigcup_{n\in\mathbb{N}}A_n)=1.$$
