Suppose we have two linear programs $Ax\leq b$ and $Bx\leq c$ is there a way to combine them into one program of possibly a larger dimension $Cy\leq d$ such that projection of vectors $y$ into a subspace of dimension $length(x)$ yields feasible points of union of original programs?


$C$ could have exponentially many constraints as $A$ and $B$ and $length(y)$ could be exponentially as big as $length(x)$. 



On OR condition in Linear Programming with exponentially many constraints
The answer is: This can be done if and only if the union is convex. 


Indeed, let $P:=\{x\colon Ax\le b\}$, $Q:=\{x\colon Bx\le c\}$, and $R:=P\cup Q$. The necessity of the convexity of $R$ was already pointed out by Robert Israel. 


Now suppose that $R$ is convex. Note that $P$ and $Q$ are convex polyhedra. Also, any extreme point $p$ of the convex set $R$ is an extreme point of either $P$ or $Q$, depending on whether $p\in P$ or $p\in Q$; similarly for the extreme rays of $R$. So, $R$ is a convex polyhedron and [hence](https://en.wikipedia.org/wiki/Convex_polytope) the intersection of a finite number of half-spaces; that is, $R=\{x\colon Cx\le d\}$ for some appropriate $C$ and $d$. 


**Added in response to a comment by the OP:** If $R$ is convex, then the plane of any face of $R$ is the plane of a face of either $P$ or $Q$, and so, the matrix $[C\ d]$ is a row-submatrix of the matrix 
$\begin{bmatrix}A&b\\B&c\end{bmatrix}$. Indeed, take any support plane $S$ of the convex polyhedron $R$. Then $S$ is a support plane of both $P$ and $Q$. Suppose now that $F:=S\cap R$ is of dimension $n-1$, where $n$ is the maximum of the dimensions of $P$ and $Q$; that is, $F$ is a face of $R$. Then either $S\cap P$ or $S\cap Q$ is of dimension $n-1$, and hence a face of $P$ or $Q$, as desired. 


