Aymptotic Convergence of Mean Estimator I want to show:

Let $x_{i}$ be an iid random variable with support $x_{i} \in [0,1]$.
Prove $n^{1/3}\frac{1}{n} \sum\limits_{i=1}^{n} (x_{i} - \mathbb{E}[x_{i}] ) \xrightarrow{p} 0$.

From what I am able to solve, I believe there are two ways to solve this problem:
1st way:
We say that a random variable converges in probability, ie,  $x_{t},\xrightarrow{p}x$ if $P(|x_{t}-x|>\epsilon)=1.$
So to show
$\theta_{n}=n^{1/3}\frac{1}{n} \sum\limits_{t=i}^{n} (x_{i} - \mathbb{E}[x_{i}] ) \xrightarrow{p} 0$
we use the definition,
$P(|\theta_{t}-\theta|>\epsilon)=P(|n^{1/3}\frac{1}{n} \sum\limits_{t=i}^{n} (x_{i} - \mathbb{E}[x_{i}] ) -0|>\epsilon)=P(|n^{1/3}(\bar{X}-\mu) |>\epsilon)$
Using Chebyshev's Inequality
$P(|n^{1/3}(\bar{X}-\mu) |>\epsilon)\leq \frac{E[n^{1/3}(\bar{X}-\mu)^2]}{\epsilon^2}=\frac{n^{1/3}E[(\bar{X}-\mu)^2]}{\epsilon^2}=\frac{n^{1/3}\sigma^2}{n\epsilon^2}=\frac{\sigma^2}{n^{2/3}\epsilon^2}\xrightarrow{n\rightarrow\infty}0.$ QED
2nd Way:
From the CLT we know
$\bar{X}=\mu+O(\frac{1}{\sqrt{n}}) \rightarrow $$  \bar{X}-\mu=O(\frac{1}{\sqrt{n}})$
$  \sqrt{n}(\bar{X}-\mu)=\sqrt{n}O(\frac{1}{\sqrt{n}})=O(1)$
which as $n\rightarrow \infty$ $O(1)\rightarrow 0$.
If instead we multiply by $n^\frac{1}{3}$
$  n^\frac{1}{3}(\bar{X}-\mu)=n^\frac{1}{3}O(\frac{1}{\sqrt{n}})=O(\frac{n^\frac{1}{3}}{\sqrt{n}})=O(\frac{1}{n^{1/6}})\xrightarrow{n\rightarrow\infty}0$. QED
My question is this:
(1) Are the two proofs I wrote valid?
(2) For the 2nd proof. I am still confused how we derived $\bar{X}-\mu=O(\frac{1}{\sqrt{n}})$. I believe it was derived using a taylor series expansion of the characteristic function, but I am not entirely certain how that derived the $O(\frac{1}{\sqrt{n}})$. Despite not knowing how to show it, I thought my logic was still correct from thereon after.
 A: For your first proof, there was a calculation error when applying the Chebyshev's inequality: $n^{1/3}$ also needs to be squared to $n^{2/3}$, fortunately, this error doesn't affect the final result.
For your second proof, the logic is correct but there is an important  conceptual mistake. Since $\bar{X}$ is random, you should write
$$\sqrt{n}(\bar{X} - \mu) = O_\color{red}{P}(1)$$
instead of $O(1)$, which is used under a deterministic/nonrandom setting (such in calculus or real analysis). By the way, $O(1)$ doesn't necessarily tend to $0$ as $n \to \infty$.
The correct writing up for your second proof can be as follows:
$$n^{1/3}(\bar{X} - \mu) = n^{-1/6}\times \sqrt{n}(\bar{X} - \mu) \Rightarrow 0$$
by central limit theorem and Slutsky's lemma. Since $0$ is a constant, it is also equivalent to converge in probability.
A: $  \bar{X}-\mu=O(\frac{1}{\sqrt{n}})$ comes from
$\bar{X}=\mu+O(\frac{1}{\sqrt{n}})$

Define $Z_n := \sqrt{n}(\frac{X_1 + ... + X_n}{n} - \mu)$
$\to \frac{Z_n}{\sqrt{n}} = (\frac{X_1 + ... + X_n}{n} - \mu)$
$\to \frac{Z_n}{\sqrt{n}} + \mu = \frac{X_1 + ... + X_n}{n}$
$\to \frac{Z_n}{\sqrt{n}} + \mu = \bar X_n$
Actually,
$$\text{your} \ 'O(\frac{1}{\sqrt{n}})' = \frac{Z_n}{\sqrt{n}}$$
Note that as you said:
$$\sqrt{n} \ 'O(\frac{1}{\sqrt{n}})' = \sqrt{n} \frac{Z_n}{\sqrt{n}}$$
$$\ 'O(1)' = Z_n$$
$$\lim_{n \to \infty} \ 'O(1)' = \lim_{n \to \infty} Z_n$$
$$= \lim_{n \to \infty} \sqrt{n}(\frac{X_1 + ... + X_n}{n} - \mu)$$
$= 0$ by CLT
