Every probability measure over countable sample space follows form of discrete measure.

Question

Context

The triple $(\Omega, \mathcal{F}, \mathbb{P})$ is a probability space with:

• Sample space $\Omega$ - which is a set,
• $\sigma$-algebra $\mathcal{F} \subseteq \mathcal{P}(\Omega)$ which is a set of subsets of $\Omega$ that has the following properties:
  1. $\Omega \in \mathcal{F}$,
  2. $A \in \mathcal{F} \implies A^c = \Omega \backslash A \in \mathcal{F}$,
  3. The sequence $(A_k)_{k = 1}^\infty \subseteq \mathcal{F} \implies \bigcup_{k = 1}^\infty A_k \in \mathcal{F}$
• Probability measure $\mathbb{P} : \mathcal{F} \to [0,\infty)$ where
  1. $\mathbb{P}(\Omega) = 1$,
  2. $\mathbb{P}\left(\bigcup_{k=1}^\infty A_k\right) = \sum_{k = 1}^\infty \mathbb{P}(A_k)$ where the $A_k$ are mutually disjoint.

An event space is a probability space without the probability measure, i.e: $(\Omega, \mathcal{F})$.

I'm considering a specific probability space $(\Omega, \mathcal{P}(\Omega), \mathbb{P})$ where $$\mathbb{P}(A) := \sum_{\omega \in A}\rho(\omega)\qquad (A \subseteq \Omega)$$ where $\rho : \Omega \to [0,1]$ is any function such that $\sum_{\omega \in \Omega}\rho(\omega) = 1$. It is not hard to prove that $\mathbb{P}$ is a probability measure. Here, the probability measure $\mathbb{P}$ is called the discrete probability measure.

The Problem

I'm trying to prove that if $\Omega$ is countable, then the probability space $(\Omega, \mathcal{P}(\Omega), \mathbb{P})$ will have its probability measure $\mathbb{P}$ always be in the form of the discrete probability measure as detailed in the context, regardless of its concrete definition.

What I've Tried

Given that $\Omega$ is countable, we can assign probabilities to all the singleton sets. This is because if we let $S$ be the collection of these singleton sets $S = \{\{\omega\}: \omega \in \Omega\}$, we have $\bigcup_{X \in S} X = \Omega$. So this implies $$1 = \mathbb{P}(\Omega) = \mathbb{P}\left(\bigcup_{X \in S} X\right) = \sum_{X \in S}\mathbb{P}(X)$$ With the last part of the equality holding because all the $X$s are mutually disjoint. However, since each $X$ simply has one element in it, we can use the axiom of choice(?) to make it such that we can define $\rho$ such that $\rho(x) = \mathbb{P}(X)$ where $x \in X$ is the only element in $X$.

From here we can see that $\sum_{\omega \in \Omega} \rho(\omega) = \sum_{X \in S}\mathbb{P}(X) = 1$. Therefore $\mathbb{P}$ is a discrete probability measure, as required.

Have I made any mistakes with my proof? If so, how can I correct them?

Thanks in advance!