Finding a maximum acyclic sub-tournament given two acyclic sub-tournaments
consider the reduction from vertex cover to the above problem. 

consider graph $G(V,E)$ with vertex V={1,2,..n}

let T be the tournament with vertices $x_{i},y_{i},z_{i}$ for vertex i=1,2...n

construct the tournament with edges in order of  $x_{1},y_{1},z_{1},x_{2},y_{2},z_{2}...x_{n},y_{n},z_{n}$.if there exist a edge(i,j) then $z_{j},x_{i}$.  $x_{i},y_{i}$ belongs to $T_{1}$ for i=1,2...n and $z_{i}$ belongs to $T_{2}$ for i=1,2...n. It is clear that $T_{1}$ and $T_{2}$ are acyclic.

Example for the construction

consider the graph G(V,E) with vertex set {1,2,3} and edges be (1,2)(2,3)
construction as follows

Tournament has vertices $x_{1},y{1},z{1},x_{2},y_{2},z_{2},x_{3},y_{3},z_{3}$
with construction as follows

Step1 
$x_{1}$ has directed edges to all vertices.

$y_{1}$ has directed edges to all vertices except $x_{1}$

$z_{1}$ has directed edges to all vertices except $x_{1},y_{1}$

$x_{1}$ has directed edges to all vertices except $x_{1},y_{1},z_{1}$

and repeat the process till all edges for the tournament constructed

step 2: (1,2) has a edge so swap the edge direction from $(x_{1},y_{2})$ to $(y_{2},x_{1})$ similarly for (1,3)
claim: G has a vertex cover of size k only if T has feedback vertex set of size k

one case is clear, if G has a vertex cover of size k then surely T has a fvs of size k

other way is, if there is edge (i,j) where $i&lt;j$ if both $x_{i},z_{j}$ is not picked into the solution then we are forced to pick all vertices $y_{i},z_{i},x_{k},y_{k},z_{k},x_{j},y_{j}$ where for all k such that $i&lt;k&lt;j$. so picking $x_{i},z_{j}$ will result in better solution.hence proved

please see and verify.