Let $P = \{x\in\mathbb{R}^n|Wx\leq b\}$ be a polytope with more than one point (strictly) such that $W\in\mathbb{R}^{m\times n}$ and $b\in\mathbb{R}^m$, and let $c\in\mathbb{R}^n$ be a vector such that $\text{max}\{c^Tx|x\in P\} - \text{min}\{c^Tx|x\in P\} = 1$.
Apparently, by 'LP-duality', we can then conclude that $(1,0,\dots, 0)\in\mathbb{R}^{n + 1}$ is in the rowspace of the matrix $[b, W]$.
I know that the dual of the max LP (above) is min$\{b^Ty|W^Ty=c, y\geq 0\}$ and the dual of the min LP (above) is max$\{b^Ty|W^Ty=c, y \leq 0\}$, but I don't see where to go from here to get the conclusion. Would someone be able to help clarify this?
Let $$A = \begin{bmatrix} b^\top \\ W^\top \end{bmatrix} = \begin{bmatrix} b & W\end{bmatrix}^\top.$$ We wish to show that $d:= (1, 0, \ldots, 0)^\top$ lies in the columnspace of $A$. In other words, there exists some $x \in \Bbb{R}^n$ such that $Ax = d$. For such $x$, we get $W^\top x = 0$ and $b^\top x = 1$.
Now, consider your two dual problems. Let $y_+$ minimise $\{b^\top y : W^\top y = c, y \ge 0\}$ and $y_-$ maximise $\{b^\top y : W^\top y = c, y \le 0\}$. Then $$W^\top(y_+ - y_-) = W^\top y_+ - W^\top y_- = c - c = 0,$$ and $b^\top(y_+ - y_-) = b^\top y_+ - b^\top y_-$ is the difference between these optimal values, i.e. $1$. Thus, $x = y_+ - y_-$ is a suitable choice for $x$.