Let's let $H=\{x\in\Bbb{R}^n: \langle x,y\rangle=b\}$ be a hyperplane in $\Bbb{R}^n$, and let $u,v\in\Bbb{R}^n$ be such that $\langle u,y\rangleb$ - in other words, $u$ and $v$ lie in opposite half-spaces. Then, let $z(t)=(1-t)u+tv,t\in[0,1]$ be the line segment joining $u$ to $v$. Now, define the function
$$
f(t)=(1-t)(\langle u, y\rangle-b)+t(\langle v, y\rangle-b)
$$
Notice that since $\langle u,y\rangleb$, we have a continuous function of $t$ such that $f(0)<0$ and $f(1)>0$. Hence by the intermediate value theorem, there exists $t_0\in [0,1]$ such that $f(t)=0$; rearranging we would have
$$
(1-t_0)(\langle u, y\rangle-b)+t_0(\langle v, y\rangle-b)=0\\
\Rightarrow (1-t_0)\langle u,y\rangle+t_0\langle v,y\rangle=b\\
\Rightarrow \langle z(t_0),y\rangle=b
$$
and hence $z(t_0)$ is on the hyperplane.