Let $(B_t)_t$ a real brownian motion. On the literature there are several proofs of the fact that $Y_t = -B_t$ is still a brownian motion.
I was trying to show myself that $B_t$ and $Y_t$ have the same law.
We know that both $B_t$ and $Y_t$ are centered Gaussian random variables, hence they have the same law iff they have the same covariance matrix $\Gamma$.
Then $$Cov(Y_t,Y_s) = E[Y_t Y_s] - 0 = E[(-B_t)(-B_s)] = E[B_t B_s] = s \wedge t$$
Now, since $Cov(B_s,B_t) = s \wedge t$ holds for any brownian motion, the covariance matrix is the same, and hence $B_t$ and $-B_t$ have the same law
Your argument is correct. The only suggestion of improvement I have is that you want to prove that the process $\left(B_t\right)_{t\geqslant 0}$ and $\left(Y_t\right)_{t\geqslant 0}$ have the same law. Therefore, you need to replace the sentence "We know that both $B_t$ and $Y_t$ are centered Gaussian random variable" by "We know that both $\left(B_t\right)_{t\geqslant 0}$ and $\left(Y_t\right)_{t\geqslant 0}$ are centered Gaussian random variable".