So an answer of this post provided an example of a sequence of random variables that converges in distribution but not in probability. There is also a post here that gives an answer on how to prove that a sequence of random variables (defined as below) does not converge in probability. I'd like to know if an alternate proof that I propose here is correct or not.
Let $U \sim \text{Unif}(0,1)$ and define
$$ X_n = \begin{cases} U, \ \ \ \ \ n \mod 2 = 0,\newline 1 - U, \ n \mod 2 = 1 \end{cases} $$ If $X \sim \text{Unif}(0,1)$. Then $X_n \to X$ in distribution since $X_n$ has the same distribution function as $X$ for any $n \in \mathbb{N}$.
If $n$ is even, then $$X_n - X = 0$$ Therefore $\mathbb{P}(|X_n - X| > \varepsilon) = 0$ since $\varepsilon > 0$. If $n$ is odd, then $$X_n - X = 1 - 2X$$ So, \begin{align*} \mathbb{P}(|X_n - X| > \varepsilon) &= \mathbb{P}(1 - 2X > \varepsilon) + \mathbb{P}(1 - 2X < - \varepsilon) \newline &= \mathbb{P}(X <(1 - \varepsilon)/2) + \mathbb{P}(X > (\varepsilon + 1)/2) \newline &= \frac{1 - \varepsilon}{2} + 1 - \frac{1 + \varepsilon}{2} = 2 \end{align*} Hence $X_n$ does not converge to $X$ in probability.
Is my reasoning correct?
This is not correct. $(X_b)$ converges in distribution to lots of r.v.'s and $X$ is not the only possible limit. To prove that this sequence does not converge in probability it is not enough to consider $X_n-X$ since $X_n$ may converge in probability to some other r.v.
$P(|X_{2n}-X_{2n-1}| >\epsilon) =P(|U-(1-U)| > \epsilon)\geq P(U>\frac {1+\epsilon} 2)$ which proves that $(X_n)$ is not Cauchy in probability. Hence, it cannot converge to any r.v. in probability.