Let $ X=(X_t)_{t \in T} $ with $ X_t: \Omega \rightarrow \mathbb{R}^n $ for all $ t \in T=[0,t_{\max}] $ a stochastic process satisfying $ X_t \stackrel{\mathcal{L}}{=} X_0 $ (e.g., a normal distribution with parameters $ \mu $ and $ \Sigma $) for all $ t \in T $.
Do we have $ P\left[ \bigcap_{t \in T} \lbrace c^\top X_t > C\rbrace \right] = P[c^\top X_0 > C] $ for all vectors $ c \in \mathbb{R}^n $? Or do we need any further assumptions?
Can we just state $ \bigcap_{t \in T} \lbrace c^\top X_t > C\rbrace = \lbrace c^\top X_0 > C \rbrace $ since the distribution of $ X_t $ does not change over time? (How would a formal proof look like?)
First, the event $\bigcap_{t \in T} \lbrace c^\top X_t > C\rbrace$ may not be measurable when $T$ is an uncountable index set. Second, your assertion need not hold even if $T=\{0,1 \}$, e.g. $X_0$ and $X_1$ are independent $N(0,1)$ so that for any $C\in\mathbb{R}$, $$ \mathsf{P}(X_0>C,X_1>C)=[\mathsf{P}(X_0>C)]^2<\mathsf{P}(X_0>C). $$