I remember a fairly lucid explanation from a while back:
The statement is true for $n=1$, since $1$ line separates the plane into $2$ regions, and $(1^2+1+2)/2=2$. Assume that inductive hypothesis, that $n$ lines of the given type separate the plane into $(n^2+n+2)/2$ regions. Consider an arrangement of $n+1$ lines. Remove the last line. Then there are $(n^2+n+2)/2$ regions by the inductive hypothesis. Now we put the last line back in, drawing it slowly, and see what happens to the regions. As we come in "from infinity," the line separates one infinite region into two (one on each side of it); this separation is complete as soon as the line hits one of the first $n$ lines. Then, as we continue drawing from this first point of intersection to the second, the line again separates one region into two. We continue in this way. Every time we come to another point of intersection between the line we are drawing and the figure already present, we lop off another additional region. Furthermore, once we leave the last point of intersection and draw our line off to infinity again, we separate another region into two. Therefore, the number of additional regions we formed is equal to the number of points of intersection plus one. Now, there are $n$ points of intersection, since our line must intersect each of the other lines in a distinct point (this is where the geometric assumptions get used). Therefore, this arrangement has $n+1$ more points of intersection than the arrangement of $n$ lines, namely $(n^2+n+2)/2+(n+1)$, which, after a bit of algebra, reduces to $[(n+1)^2+(n+1)+2]/2$, exactly as desired.