There is a purely probability theoretical argument in the proof of Lévy's characterization of Brownian motion, which I do not completely understand. I think it is rather easy. Suppose we know $E[e^{iu^{tr}(X_t-X_s)}|\mathcal{F}_s]=e^{-\frac{1}{2}|u|^2(t-s)}$
for all $u\in\mathbb{R}^d$. From this it should follow, that $X_t-X_s$ is independent of $\mathcal{F}_s$ and normally distributed with mean $0$ and covariance matrix $(t-s)Id_{d\times d}$, hence the $X^k$ should be independent Brownian Motion ($^k$ denotes the k-th coordinate). My suggestion is to take expectation:
$E[e^{iu^{tr}(X_t-X_s)}]=e^{-\frac{1}{2}|u|^2(t-s)}$
Hence I know that $(X_t-X_s)$ has the right distribution. Furthermore, by the structure of the convariance matrix, I know $(X^i_t-X_s^i)(X^k_t-X^k_t)$ are uncorrelated for $k\not=i$. Why should the $X^k$ be independent. I have independence of the product of increments, how do I get independence for $X^k$? I guess, this uses, that every coordinate is a normal distributed r.v. and for normal distributed r.v. "uncorrelated implies independent". Even more, I do not see how independence of $\mathcal{F}_s$ should follow. So any help would be appreciated. Thanks in advance!
math