I know that to prove this I have to shown that a set of finite inequalities make a line in $\mathbb R^n$ that is
$$ L = \{ x_0 + \lambda d : \lambda \in \mathbb R^n \} $$
But can we say a line is the intersection of 3 hyperplanes in $\mathbb R^n$ (since each hyperplane can be shown with 2 inequalities). Can we say the same things about a half-line (that is when $ \lambda \geq 0 $)?
I am assuming that $d \neq 0$ below:
Let $v_2,...,v_n$ form a basis for $(\operatorname{sp}\{d\})^\bot$.
Define $H_k = \{x| \langle v_k, x-x_0 \rangle = 0 \}$.
Then $L= \cap_k H_k$.
Suppose $x \in L$, then $x=x_0+\lambda d$, and it is easy to check that $x \in H_k$ for all $k$.
In general, given any $x$, there are unique $\lambda, \alpha_k$ such that $x=x_0+\lambda d + \sum_k \alpha_k v_k$.
Suppose $x \in H_k$, then we must have $\alpha_k = 0$.
Hence if $x \in \cap_k H_k$, then $x=x_0 + \lambda d$ and so $x \in L$.