Proving/disproving that differences $X_2 - X_1, \ldots, X_n - X_{n-1}$ of independent RVs $X_1, \ldots, X_n$ are also independent. Let $X_1,\ldots,X_n$ be pairwise independent RVs in a Euclidean space.  My question is:

Is the set of differences $\{X_{j+1} - X_j \}_{j=1}^{n-1}$ also pairwise independent?

(Edited: by the discussion, it turns out that the proof below is wrong--see the answers)
If they are all Gaussian, then 
\begin{equation}
    X_j, \, X_k \textrm{ independent } \; \Longleftrightarrow \; X_j, X_k \textrm{ uncorrelated } \; \Longleftrightarrow \; \mathbb{E}[X_jX_k] = \mathbb{E} [X_j] \mathbb{E} [X_k], 
\end{equation}
so the above statement can be proven as follows.

Let $\mu_i = \mathbb{E}[X_i]$ for  $i = 1,2,\ldots,n$. Then, since $X_{j-1}$, $X_j$, and $X_{j+1}$ are all uncorrelated, 
\begin{align}
\mathbb{E}\big [ \, (X_j - X_{j-1}) \cdot (X_{j+1} - X_j) \, \big ] 
  &= \mu_j \mu_{j+1} - \mu_j^2 - \mu_{j-1}\mu_{j+1} + \mu_{j-1} \mu_j \\
  &= (\mu_j - \mu_{j-1}) \cdot (\mu_{j+1} - \mu_j ) \\
  &= \mathbb{E}[X_j - X_{j-1}]\cdot \mathbb{E}[X_{j+1} - X_j].
\end{align}
Hence, $X_j - X_{j-1}$ and $X_{j+1} - X_j$ are also uncorrelated. Since $X_j$'s are all Gaussian, so are $X_{j+1} - X_{j}$ for all $j=1,2,\ldots,n-1$. Therefore,  $\{X_{j+1} - X_j \}_{j=1}^{n-1}$ is independent. 

(Edited: The statement below regarding non-Gaussian RVs is also not true--see the answer)
In general non-Gaussian situations, the same proof-line shows that $\{X_{j+1} - X_j \}_{j=1}^{n-1}$ is uncorrelated, but not necessarily independent since independent RVs are always uncorrelated, but not vice versa in general. 
I wonder if there is any other specific non-Gaussian distributions that make the statement true? A counter example? Or, is there any general proof of the statement w/o assuming RVs are Gaussian?
Many thanks in advance for your discussion and comments.

Let $X_1, X_2, X_3$ be independent identical random variables uniformly distributed on $[0,1]$. We construct random variables $Y_1=X_2-X_1$, $Y_2=X_3-X_2$.
If I tell you that $Y_1=-1$, then $X_2=0$, so $Y_2\geq 0$, consequently $Y_1$ and $Y_2$ are dependent.

PS: Your derivation is wrong:
\begin{align}
\mathbb{E}\big [(X_j - X_{j-1}) \cdot (X_{j+1} - X_j)] 
  &= \mathbb{E}[X_j X_{j+1}-X_j^2-X_{j-1}X_{j+1}+X_{j-1}X_j] \\
  &= \mu_j \mu_{j+1}-\mathbb{E}[X_j^2]-\mu_{j-1}\mu_{j+1} + \mu_{j-1} \mu_j \\
  &= \mu_j \mu_{j+1}-\mu_j^2-\mathbb{var}[X_j^2]-\mu_{j-1}\mu_{j+1} + \mu_{j-1} \mu_j \\
  &\neq \mu_j \mu_{j+1} - \mu_j^2 - \mu_{j-1}\mu_{j+1} + \mu_{j-1} \mu_j \\
  &= (\mu_j - \mu_{j-1}) \cdot (\mu_{j+1} - \mu_j ) \\
  &= \mathbb{E}[X_j - X_{j-1}]\cdot \mathbb{E}[X_{j+1} - X_j].
\end{align}
so the assumption is wrong for Gaussians too. Actually differences of independent random variables are never independent, except for the case when some middle r.v's are constants.
