I need a check on the following exercise. In particular, I'd like to be sure that points ii), iii) are okay.
Let $(B_t)_{t \in [0,1]}$ be a standard Brownian motion. Prove that the process $W_t=B(1-t) - B(1)$ defined for $t \in [0,1]$ is a standard Brownian motion
i) First I can see that $W(0)=B(1) - B(1)=0$ a.s.
ii) I need to check the the increments are independent. To this aim, I fix a partition of $[0,1]$, $0<t_1 < t_2 < \ldots < t_n$ and consider the increments $(W_{t_1}, W_{t_2} - W_{t_1}, \ldots, W_{t_n} - W_{t_{n-1}})$
I have that $$W_{t_{i}} - W_{t_{i-1}} = B(1 - t_i) - B(1 - t_{i-1})$$
and $$W(t_{i+1}) - W(t_i) = B(1 - t_{i+1}) - B(1-t_i)$$
I need to check they're independent: since both of them are normally distributed, to check the independence is enough to check that the covariance is $0$. Moreover, since $B$ is a Brownian motion, I can use the fact that
$Cov(B_t, B_s)= s \wedge t$.
Hence $$Cov(B(1-t_i) - B(1-t_{i+1} ), B(1-t_{i-1}) - B(1-t_i)) = \\ (1-t_i) \wedge (1-t_{i-1}) - (1-t_{i+1}) \wedge (1-t_{i-1}) - (1-t_i) \wedge (1-t_i) + (1-t_{i+1}) \wedge (1- t_i) = 0$$
Hence the increments are independent.
iii) I need to show that for evert $0\leq s <t$: $W(t) - W(s) - N(0,t-s)$
I have that $W(t) - W(s) = B(1-t) - B(1-s)$. Now I have that $1-t< 1 -s$, therefore I note that $$ B(1-t) - B(1-s) = -(B(1-s) - B(1-t)) $$
Now I see that the right hand side has normal law since it's an increment of the Brownian motion. Moreover, I have that it's mean is clearly $0$, while the variance is $t-s$
Therefore $W(t) - W(s) - N(0, t-s)$.
(iv) Path continuity
I have that $t \mapsto B_t(\omega)$ is continuous for $P$-almost every $\omega$. Hence, the function $$t \mapsto B_{1-t}(\omega) - B(1)$$ is continuous, since is obtained from $B_t$ by a traslation and adding a constant $B(1)$.
Edit
To show it's a Gaussian process, I can note the following:
first I note that $(B(1-t_n), B(1-t_{n-1}), \ldots, B(1-t_1), B(1))$ is a Gaussian vector (since $B$ is a Brownian motion). Therefore, I can see
$\begin{pmatrix} W(t_1) \\ \vdots \\ W(t_n) \end{pmatrix}=T((B(1-t_n), B(1-t_{n-1}), \ldots, B(1-t_1), B(1)))$
where $T$ is the linear map $T(x_1,x_2, \ldots,x_n)=(x_{n-1} - x_n,x_{n-2} - x_n, \ldots, x_1 - x_n )$.
Hence, since it's the image via a linear map of a Gaussian vector, $\begin{pmatrix} W(t_1) \\ \vdots \\ W(t_n) \end{pmatrix}$ is a Gaussian vector.
Your proof looks good for the most part. However, it should be noted that two random variables being normal and having zero covariance does not imply independence. Wikipedia even has a nicely titled page for this.
Your calculation can be justified, however. The key point is that the standard Brownian motion taken at any two times $s,t$ produces two jointly Gaussian variables. It suffices to show that two jointly Gaussian variables are uncorrelated to show independence.
EDIT: Justification of independent increments.
Consider any two disjoint increments $[s_1, t_1]$ and $[s_2, t_2]$ of the process $(X)_t = B_{1-t} - B_1$. We want to show that $Y_1 = X_{t_1} - X_{s_1} = B_{1-t_1} - B_{1-s_1} $ is independent of $Y_2 = X_{t_2} - X_{s_1} = B_{1-t_2} - B_{1-s_2} $. It's actually easier to forego the covariance calculation. $Y_1 = B_{1-t_1} - B_{1-s_1}$ is the random variable tracking the Brownian motion in the time period $[1 - t_1, 1 - s_1]$ (we've flipped the endpoints as the negative of a standard Brownian motion is also a standard Brownian motion). Similarly, $Y_2 = B_{1-t_2} - B_{1-s_2}$ is the random variable tracking the Brownian motion in the time period $[1 - t_2, 1 - s_2]$. Since $[s_1, t_1]$ and $[s_2, t_2]$ are disjoint by assumption, the intervals $[1 - t_1, 1 - s_1]$ and $[1 - t_2, 1 - s_2]$ are also disjoint, and hence by the independent increments property of the standard Brownian motion, $Y_1 = B_{1-t_1} - B_{1-s_1}$ and $Y_2 = B_{1-t_2} - B_{1-s_2}$ must be independent too.