Denote $M$ as a $m\times n$ matrix whose components are all nonnegative integers (actually 0 or 1) and $1$ as the $m$ dimensional vector $(1,1,\cdots,1)$. Show that:
There is a vector $x_0$ minimize $1^Tx$ such that $Mx \geq 1$ where the components of $x$ are nonnegative, and the components of $x_0$ are all rational numbers.
This question is from a sentence in the bottom of page 3 of a book,
We may assume that $x$ is rational (since the data in the LP are all integers).
In general your theorem doesn't hold, because there is the possibility that $Mx\geq1$ do not upperbound your function $1^Tx $.
Example in $\mathbb{R}^2$:
$-x_1-x_2\geq 1$
There is no vector $x_0\in \mathbb{R}^2$, which minimize your function since your function $x_1+x_2$ always decreases with $x_1$, $x_2$ going into direction $-\infty$.
Such problems like yours are often bounded by additional sign constraints:
$x\geq 0 $
With your function $1^Tx$ and this additional constraint your theorem holds, because $1^Tx$ is upperbounded by $x=0$ and then your optimal point $x_0$ is a corner (or an arbitrary point on a whole facet) of the konvex set $Mx\geq1,\;x\geq0$.
(the minimum $x_0$ is rational, because your matrix M consists of integers and a corner of $Mx\geq1,\; x\geq0$ is a solution of a linear equation system over a subset of the equations $Mx=1, x=0$.)