Let $W_t = (W_t^{(1)},W_t^{(2)},W_t^{(3)})$ be 3 dimensional Brownian motion. Let $X=sgn(W_1^{(1)})sgn(W_1^{(2)})sgn(W_1^{(3)})$. Define a 3 dimensional process $M_t$ as follows :
$M_t^{(1)} = W_t^{(1)}, M_t^{(2)} = W_t^{(2)}, M_t^{(3)} = X W_t^{(3)}$
Prove that $(M_t^{(1)},M_t^{(3)})$ is 2 dimensional Brownian motion but $(M_t^{(1)},M_t^{(2)},M_t^{(3)})$ is not a Brownian motion.
To show that $(M^{(1)},M^{(3)})$ is a Brownian motion, note that $(M^{(1)},M^{(3)})=(W^{(1)},XW^{(3)})$ where $X=\pm1$ is independent of $(W^{(1)},W^{(3)})$, hence the distribution of $(M^{(1)},M^{(3)})$ is a barycenter of those of $(W^{(1)},W^{(3)})$ and $(W^{(1)},-W^{(3)})$. Since these last two distributions coincide and $(W^{(1)},W^{(3)})$ is a Brownian motion, the proof is complete.
To show that $(M^{(1)},M^{(2)},M^{(3)})$ is not a Brownian motion, note that $M^{(1)}_1M^{(2)}_1M^{(3)}_1=|W^{(1)}_1W^{(2)}_1W^{(3)}_1|$ hence the distribution of $(M^{(1)}_1,M^{(2)}_1,M^{(3)}_1)$ allocates measure zero to the set $\{(x,y,z)\in\mathbb R^3\mid xyz\lt0\}$ while the distribution of 3-dimensional Brownian motion at time $1$ does not. This argument shows that $(M^{(1)}_1,M^{(2)}_1,M^{(3)}_1)$ is not even gaussian.