In a tree let the routes between vertices $a\to b$ and and $c\to d$ be vertex-disjoint. Show that $a\to c$ and $b \to d$ have vertices in common. Let the route between $u$ and $v$ in a simple, connected graph be the shortest sequence of vertices, such that $[u,u_1,...,u_n,v]$ would be a way to travel a graph from vertex $u$ to $v$.
In a tree let the routes between vertices  $a\to b$ and and $c\to d$ be vertex-disjoint. Show that $a\to c$ and $b \to d$ have vertices in common.
Here's the sketch of my attempt.
Let us assume to the contrary that $a\to c$ and $b \to d$ are also vertex disjoint. 
Let us inspect routes $a \to b$ and $a \to c$. They can have some vertices in common, but they must run out at some point, otherwise $a \to c$ and $c \to d$ would cross, so the route must look like this:
Meaning that for a while the routes are the same, but then they diverge. The let us think about $c \to d$ and $a \to c$, they too can have a vertex is common, but becuase $c \to d$ and $a \to b$ are disjoint it has to fall "below" $A'$, meaning that it look a little like this:

Similar reasoning leads us to finding $B'$ and $D'$, meaning that we will have four vertices $A',B',C',D'$ that will be in a cycle, meaning that the graph can't be a tree.
Is my reasoning correct?
 A: Your reasoning is essentially correct.  If you want to make it more precise (so that you don't need to say things like " 'below' $A^\prime$ " or rely on possibly misleading pictures), you can change the proof a bit.  What follows is not really different than your reasoning, but merely a different and perhaps more precise way of expressing it.
First, argue (as you did) that there is a vertex $A^\prime$ where the paths $a\rightarrow c$ and $a\rightarrow b$ diverge.  Then argue that the vertices $A^\prime$, $b$, $c$, and $d$ satisfy the same conditions on vertex-disjointness as $a, b, c, d$, because the paths are strictly contained in the original paths.  Additionally, $A^\prime\rightarrow b$ and $A^\prime \rightarrow c$ have only the vertex $A^\prime$ in common by choice of $A^\prime$.
Now, instead of having to draw more and more complicated pictures, you can actually apply the exact same argument to $A^\prime$, $b,c,d$ with $b$ as the vertex of interest instead of $a$, getting $B^\prime$, and so on.
Applying this a total of four times, one then has that $A^\prime, B^\prime, C^\prime, D^\prime$ have the following properties:  $A^\prime\rightarrow B^\prime$ and $C^\prime \rightarrow D^\prime$ are vertex-disjoint, $A^\prime\rightarrow C^\prime$ and $B^\prime \rightarrow D^\prime$ are vertex-disjoint, and each pair of consecutive paths (e.g. $A^\prime \rightarrow B^\prime$ and $B^\prime \rightarrow C^\prime$) have only one vertex in common.  It's then easy to verify that $A^\prime\rightarrow B^\prime \rightarrow C^\prime \rightarrow D^\prime \rightarrow A^\prime$ is in fact a cycle.
