Non-atomic finite measure space $(X,\mathcal{A})$ allows a partition $\left( X_i\right)_{i=1}^m$ Let $(X,\mathcal{A},\mu)$ be a finite and non-atomic measure space. I want to show that for every $\varepsilon >0$ there is a finite partition of $X$ into subsets $X_1,...,X_m$ with $\mu(X_i)<\varepsilon$.
I think that I have found a simple proof, but have doubts because it is so simple.
Proof: 
By the property of non-atomic I can find a sequence $X\supset A_1^1\supset A_2^1\supset ... \supset A_n^1\supset ... $ with 
$\mu(X)> \mu(A_1^1)> \mu(A_2^1)> ... > \mu(A_n^1)> ...>0$ where $\mu(A_n^1)<\varepsilon$
Now we do the same for $X\setminus A_n^1$: 
$X\setminus A_n^1\supset A_1^2\supset A_2^2\supset ... \supset A_n^2\supset ... $ with 
$\mu(X\setminus A_n^1)> \mu(A_1^2)> \mu(A_2^2)> ... > \mu(A_n^2)> ...>0$ where $\mu(A_n^2)<\varepsilon$
Then we just repeat with $X\setminus (A_n^1\cup A_n^2)$. Since the measure space is finite, there must be (use monotonicity of the measure) an $m\in\mathbb{N}$ with $X=\bigcup_{i=1}^m A_n^i$ where the $A_n^i$ are disjoint by construction.
 A: This is immediate from


Theorem (Seirpinski). If $\mu$  is a non-atomic measure on $X$ and $0<\alpha<\mu(X)$ then there exists $E\subset X$ with $\mu(E)=\alpha$.


Incidentally, it seems to me that the comments regarding what's wrong with your proof  of the existence of that partition are close to backwards. First, although you didn't explain why, which you need to  do, getting  $\mu(A)<\epsilon$ is easy; perhaps "trivial",  certainly not "highly non-trivial":


Easy Lemma. If $\mu$ is a non-trivial finite non-atomic measure on $X$ then there exists $E\subset X$ with $0<\mu(E)\le\mu(X)/2$.


Proof: Since $\mu$ is non-atomic there exists $E_1\subset X$ with $0<\mu(E_1)<\mu(X)$. Now either $E=E_1$ or $E=X\setminus E_1$ works.
Applying that finitely many times gives $0<\mu(A)<\epsilon$, no problem. The actual problem with your construction is this:
We start by finding  $E_1\subset X$ with $0<\mu(E_1)<\epsilon$, then $E_2\subset X\setminus E_1$ with $0<\mu(E_2)<\epsilon$, etc. It does not follow that there exists $n$ with $X=E_1\cup\dots\cup E_n$ (modulo null sets). 
The problem is that $\mu(E_k)$ may be too small; for example if $\mu(E_k)=\mu(X)/3^k$ then $$\mu\left(X\setminus\bigcup_{k=1}^\infty E_k\right)>0.$$
(Indeed, this is exactly why Seirpinski's Theorem is not a trivial consequence of the Easy Lemma; note that the theorem does follow in a very straightforward way from an improved version:


Somehwhat Harder Lemma. If $\mu$ is a non-trivial finite non-atomic measure on $X$ then there exists $E\subset X$ with $\frac13\mu(X)\le\mu(E)\le\frac23\mu(X)$.)


