I'll tell you the proof that's "easy and straightforward" to me, because I like toric varieties, $X$ is toric, and $dH-kE$ is an equivariant line bundle.
On a projective toric variety, an ample equivariant line bundle gives you a lattice polytope whose fan is that of the toric variety. For $\mathbb P^3$ the polytope is a tetrahedron, and $\mathcal O(d)$ gets you a tetrahedron with edge-length $d$. For $X$ you snip off one corner, obtaining something with two parallel triangular faces. If $dH-kE$ is ample, the smaller triangle has edge-length $k$, and the distance between the two triangles is $d-k$.
You want $k=1$ apparently, i.e. you just barely snip off the corner. If you took $d=1$ the snipped tetrahedron would degenerate to a triangle, and the line bundle would be only nef, not ample; the map to projective space would be the one collapsing $X$ to $\mathbb P^2$.
So you need $d\geq 2$ for this to be ample. I don't know where that $5$ comes from.
(It's fun to consider $k$ negative, where instead of the corner moving into the tetrahedron and becoming a triangle, it comes out as a sort of negative triangle, and the polytope is the size $d$ tetrahedron plus an extra size $-k$ "negative" tetrahedron. The extra one is telling you something about $H^1$ of this non-ample line bundle.)