Let $(\Omega, \mathscr{F}, \mathbb{P})$ be a probability space. Here is a hint for the exercise to prove that a stochastic process $X: [0,\infty)\times \Omega \rightarrow \mathbb{R}$ which has independent increments and $X_0=0$ is a Markov process (i.e. $\mathbb{P}(B|\mathscr{F_{=t}})= \mathbb{P}(B|\mathscr{F_{\leq t}}) $ for all $B\in \mathscr{F}_{\geq t}$):
Fix $t\in [0,\infty)$. Prove that $$\mathbb{P}(B|\mathscr{F_{=t}})= \mathbb{P}(B|\mathscr{F_{\leq t}}) $$ holds when $B$ has the form $$B=\{X_t\in E_0, ~~X_{t_1}-X_{t}\in E_1,..., X_{t_k}-X_{t_{k-1}}\in E_k\} $$ where the $E_i$'s are Borel sets and $t<t_1<...<t_k$. Then prove that the set of $B$ for which $\mathbb{P}(B|\mathscr{F_{=t}})= \mathbb{P}(B|\mathscr{F_{\leq t}}) $ is a $\sigma$-algebra.
It's not hard to do the first part. But why should the set of $B$ for which $\mathbb{P}(B|\mathscr{F_{=t}})= \mathbb{P}(B|\mathscr{F_{\leq t}}) $ form a $\sigma$-algebra? In particular, it's clear that the set of such $B$ is a $\lambda$-system, and probably also a monotone class. But why should it be closed under intersections?
Following @JasonSwanson's suggestion:
Fix $T= \{t_1,t_2,...,t_k\}$. Then the set of $B\in \mathscr{F}$ having the form $$B=\{X_t\in E_0, ~~X_{t_1}-X_{t}\in E_1,..., X_{t_k}-X_{t_{k-1}}\in E_k\} $$ is closed under finite intersection. Thus by the $\pi-\lambda$ theorem, all elements $B$ of the $\sigma$-algebra generated by $X_t, ~~X_{t_1}-X_{t},..., X_{t_k}-X_{t_{k-1}}$ satisfy $$\mathbb{P}(B|\mathscr{F_{=t}})= \mathbb{P}(B|\mathscr{F_{\leq t}}) .$$ But this $\sigma$-algebra is the same as $\mathscr{F}_{t,t_1,...,t_k}:=\mathscr{F}_T$, the $\sigma$-algebra generated by $X_t,X_{t_1},...,X_{t_k}.$ Now the union $$\bigcup_{\{T:~~ t<t_1<t_2<...<t_k, ~~k\in \mathbb{N}\}}\mathscr{F}_T$$ is in fact also a $\pi$ system (in fact an algebra), since if $F_1\in \mathscr{F}_{T_1}, F_2\in \mathscr{F}_{T_2} $, then $$F_1\cap F_2 \in \mathscr{F}_{T_1\cup T_2}.$$ Thus by the $\pi-\lambda$ theorem again, all elements $B$ of $\mathscr{F}_{\geq t}$ satisfy $$\mathbb{P}(B|\mathscr{F_{=t}})= \mathbb{P}(B|\mathscr{F_{\leq t}}) .$$