Show there is only one trajectory passing through each point I have to show the following: 
Let $\varphi$ be a flow on the manifold M and suppose that that the orbits {$\varphi_t (x_0)$} & {$\varphi_t (x_1)$} intersect. Prove that the orbits coincide.
 A: $\varphi$ being a flow means that we have a family $(\varphi_t)_{t\in \mathbb R}$ of diffeomorphisms $\varphi_t:M\rightarrow M$ indexed by $t \in \mathbb R$ which satisfies the "group law", that is $\varphi_0 = id_M$ and $\forall t,s\in \mathbb R: \varphi_t \circ \varphi_s = \varphi_{t+s}.$ This implies in particular $\forall t \in \mathbb R: (\varphi_t)^{-1} = \varphi_{-t}.$ More conceptually, the map $t\mapsto \varphi_t$ is a homomorphism from the group $(\mathbb R,+)$ to $Diffeo(M),$ the group of diffeomorphisms of $M$.
Now, the orbits $\{\varphi_t(x_0)\}$ and $\{ \varphi_t(x_1)\}$ intersect iff there are $t,s \in \mathbb R$ such that $\varphi_t(x_0) = \varphi_s(x_1).$ Applying $\varphi_{-s},$ we conclude
$$
x_1 = \varphi_0(x_1) = \varphi_{s-s}(x_1) = \varphi_{-s}(\varphi_s(x_1)) = \varphi_{-s}(\varphi_t(x_0)) = \varphi_{t-s}(x_0).
$$
Applying $\varphi_u$ for any $u\in \mathbb R$ to this, we get
$$
\forall u \in \mathbb R: \varphi_u(x_1) = \varphi_u(\varphi_{t-s}(x_0)) = \varphi_{u+t-s}(x_0).
$$
But as $u$ ranges over all of $\mathbb R$, $u+t-s$ ranges over all of $\mathbb R$ as well. So we have
$$
\{\varphi_v(x_1)\ | \ v \in \mathbb R\} = \{\varphi_v(x_0)\ | \ v \in \mathbb R\},
$$
as desired.
