If $z\in E$ is light-like, then
$$
g(x,y)\colon=\la x,y\ra+\la x,z\ra\la y,z\ra
$$
is again a Lorentz metric on $E$.
Suppose there is some $y\in E\sm\{0\}$, such that for all $x\in E$:
$$
0=g(y,x)=\la y,x\ra+\la x,z\ra\la y,z\ra~.
$$
Put $x=z$, then: $\la y,z\ra=0$ and for $x=y$ we get: $0=\la y,y\ra+\la y,z\ra^2$. It follows that $y$ is light-like and orthogonal to $z$; by exam $y=\l z$. Since $g(x,z)=\la x,z\ra$, $g$ is non-degenerated.
Next we prove that $g$ has index $1$: For $\la u,u\ra=1$, $\la v,v\ra=-1$, $v\perp u$ and $z=u+v$ we put $y=sv+tu$ an conclude that
$$
g(y,y)
=\la y,y\ra+\la y,z\ra^2
=-s^2+t^2+(-s+t)^2
=2t^2-2ts~.
$$
$g$ is a quadratic form on $\lhull{v,u}$ with eigen values $1\pm\sqrt2$, i.e. $g$ is a Lorentz metric on $\lhull{v,u}$. Finally, on $\lhull{v,u}^\perp$ $g$ coincides with $\la.,.\ra$ and since the latter is euclidean on $\lhull{v,u}^\perp$, $g$ must have index $1$.