If $\{X_t\}_{t\in [0,1]}$ is a one-dimensional Brownian motion with $X_0 = 0$ a.s. and $0\leq t_1<t_2<\cdots <t_n \leq 1$. Does it then hold that $X_{t_1},\dotsc, X_{t_n}$ are independent, and if so what is the intuition behind it?
I would say that they are independent based on the following argument but maybe there is something wrong with it and would be happy to be corrected in that case.
Let $A = \begin{pmatrix}1 & 0 & \cdots & 0\\ 1 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1\end{pmatrix}$, $Y = (X_{t_1}, X_{t_2}-X_{t_1},\dotsc, X_{t_n}-X_{t_{n-1}})$ and $X = (X_{t_1},X_{t_2},\dotsc,X_{t_n})$. Then $AY = X$ and hence it suffices to show that the distribution of $AY$: $\mathbb{P}_{AY}$ is a product measure on $\mathbb{R}^n$ to conclude that the components of $X$ are independent.
This however follows from the following computation: Let $B\in \mathscr{B}(\mathbb{R}^n)$ (Borel set) then $$\mathbb{P}(AY\in B) = \int_{\mathbb{R}^n}1_{B}(x)\mathrm{d}\mathbb{P}_{Y}A^{-1} = \int_{\mathbb{R}^n}1_{B}(Ax)\mathrm{d}\mathbb{P}_{Y} = \int_{\mathbb{R}^n}1_{B}(x)\underbrace{|\det(A^{-1})|}_{=1}\mathrm{d}\mathbb{P}_{Y} = \mathbb{P}(Y\in B).$$ Since $\mathbb{P}_Y$ is a product measure this concludes the argument.
$X_t$ and $X_s$ are not independent for $0<t<s$: $Cov (X_t,X_s)=EX_tX_s=\min \{s,t\} \neq 0$.