proof verification - Counting measure $sigma$-finite / not $sigma$-finite for different sets

Let $\zeta\colon\mathcal{P}(\Omega)\to [0,\infty]$ be the counting measure. Show that for $\Omega:=\mathbb{R}$ it is not $\sigma$-finite but for $\Omega:=\mathbb{N}$.

Hello and good afternoon! Here are my tries:

(1) Consider $\Omega:=\mathbb{N}$.

Define $A_n:=\left\{1,\ldots,n\right\}, n\geq 2$. Then $A_n\in\mathcal{P}(\mathbb{N}), A_n\nearrow\mathbb{N}$ and $\zeta(A_n)=n<\infty$ for $n\geq 1$. So the measure $\zeta$ is $\sigma$-finite.

(2) Now $\Omega:=\mathbb{R}$. First of all it is $\zeta(\mathbb{R})=\infty$ because the counting
measure of all sets that are not finite, is $\infty$.

Assume there are $A_i\in\mathcal{P}(\mathbb{R}), i\geq 1, A_i\nearrow \mathbb{R}$ and $\zeta(A_i)<\infty, i\geq 1$. Then for $A:=\bigcup_{i\geq 1}A_i$ it is because of the $\sigma$-additivity of $\zeta$ and the subtractivity
$$\zeta(A)=A_1\uplus\biguplus_{k\geq 2}(A_k\setminus A_{k-1})=\zeta(A_1)+\sum_{k\geq 2}\zeta(A_k\setminus A_{k-1})=\zeta(A_1)+\sum_{k\geq 2}\zeta(A_k)-\zeta(A_{k-1})=\zeta(A_{\infty})<\infty,$$
so because of the continuity of the measure it is

$$\zeta(A_i)\nearrow\zeta(A)<\infty.$$
On the other hand it is $\zeta(A_i)\nearrow\zeta(\mathbb{R})=\infty$. So $\mathbb{R}\neq A$.

Contradiction: It was assumed that $A_i\nearrow\mathbb{R}$, i.e. $A_i\nearrow A=\mathbb{R}$.

So there do not exist $A_i\in\mathcal{P}(\mathbb{R}), i\geq 1$ with $A_i\nearrow\mathbb{R}$ and $\zeta(A_i)<\infty$, i.e. $\zeta$ is not $\sigma$-finite for $\Omega:=\mathbb{R}$.

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Maybe you can say me, if I am right.

Miro