This is not the usual request for an intuitive proof, which has been asked already.
Having looked at various sources, I've basically concluded that Heron's formula relies on proving $$xyz = x+y+z$$ where $x$, $y$ and $z$ are the lengths between the meeting point of the incircle and the sides, and the vertices. If I take the diagram below, $x$ is $CX$, and $z$ is $XA$. $y$ would be the segment from $B$ to the incircle meeting the side, marked by a black dot, which is also $YC$. Hence $x+y+z=s$ where $s$ is the semiperimeter. From here it is easy to show that if the radius ($r$) of the incircle is $1$, then the area of the triangle is $x+y+z$. The radius does not really matter, because when it comes to proving the formula, you can just proportionally reduce the size of the triangle by a factor of $1/r^2$. So we can work with the $r=1$ case.
Proving $xyz = x+y+z$ visually is not difficult by looking at this diagram: http://jwilson.coe.uga.edu/emt725/Heron/Heron2/Heron2.html
When $r = 1$, then $EY = xy$ ($x = (s-c)$ and $y = (s-b)$ since $s = x+y+z$ as defined above) and $EY$ is also equal to $s/z$, so it is conceptually not difficult to show that $x+y+z = xyz$ and from this that $A^2 = (x+y+z)xyz = s(s-a)(s-b)(s-c)$.
But to my mind, it would be even better if we could show that $xyz$ corresponds to the area of the triangle directly, rather than messing about with equivalences. If we take the $\Delta AYE$ triangle in the above diagram and form a rectangle, whose new vertex we call $P$, and we extend a line from point $X$ to the side $EP$, calling this new vertex $Q$, then $PQXA$ is meant to be of the same area as the triangle, since $XA=z$ and $EY=xy$. Any ways to prove that rectangle $PQXA$ is equivalent to triangle $\Delta ABC$? Or perhaps there's a better way to prove $xyz$ corresponds to the area of the triangle $\Delta ABC$?