Key fact
For any
$$ 0are independent normal random variables.
Consequence
The vector
$$ (B_{t_1},\dots,B_{t_k}) $$is multivariate normal for every finite set of times. Therefore Brownian motion is a Gaussian process.
Covariance
For $s\le t$,
$$ \mathrm{Cov}(B_s,B_t)=s. $$Equivalently,
$$ \mathrm{Cov}(B_s,B_t)=\min(s,t). $$