Here is another excellent solution from
Andrei of Tudor Vianu National College, Bucharest,
Romania.
I shall prove Pick's Theorem by proving that Pick's function
$$F(P)= {\rm area }(P)  (i + {p\over 2}  1)$$ is 0 for any planar
polygon, following the steps from the problem.
(1) Let $P_1$ and $P_2$ be two polygons with a common edge, $P$ the
union of $P_1$ and $P_2$, $i$, $i_1$, $i_2$ the lattice points
inside the 3 polygons, and $p$, $p_1$, $p_2$ the lattice points on
the perimeters of the 3 polygons.
Let $x$ be the number of lattice points of the common edge of $P_1$
and $P_2$. It is clear that area($P$) = area($P_1$) +
area($P_2$)

From the figure I observe that:
$i = i_1 + i_2 + x  2$
$p = p_1 + p_2  2x + 2.$
Now, I shall calculate $F(P)$:

\begin{eqnarray} F(P)&=& {\rm area }(P)  (i + {p\over 2} 
1)\\ &=& {\rm area }(P_1)+{\rm area }(P_2)i_1 i_2 x + 2
 {p_1 \over 2}  {p_2\over 2} + x  1 + 1 \\ &=& {\rm area
}(P_1)+{\rm area }(P_2)i_1 i_2  {p_1\over 2}  {p_2\over 2} +
2\\ &=& {\rm area }(P_1)  (i + {p_1\over 2}  1)+{\rm area
}(P_2) (i + {p_2\over 2}  1)\\ &=& F(P_1) + F(P_2)
\end{eqnarray} So, if Pick's function is zero for any two of these
polygons it must be zero for the third. This means that if Pick's
Theorem holds for any two of these polygons it must hold for the
third. As a generalization, $$F(P_1+P_2+ ... +P_n) = F(P_1) +
F(P_2) + ... +F(P_n).$$ So, $F$ is additive.
(2) Now I shall prove Pick's Theorem for a rectangle with vertices
$(0,0), (a,0), (a,b), (0,b)$.
${\rm Area}(P) = ab$, $i = (a  1)(b  1)$ and $p = 2(a + b).$ So
$$F(P) = ab  (a1)(b1)  (a+b) + 1 = 0.$$ Any rectangle with
sides parallel to Ox and Oy could be translated to a rectangle of
the given form.
(3) If the rectangle is split into two triangles by a diameter then
each triangle has the same area and the same number of interior
points and the same number of points on the perimeter and hence
Pick's function must take the same value for each triangle. By
parts (1) and (2) Pick's function must be zero for such triangles.
When a polygon with vertices at lattice points is translated so
that the image has vertices at lattice points there is no change in
area, or in the number of interior lattice points, or in the number
of lattice points on the perimeter. Hence Pick's function is zero
for all rectangles and for any rightangled triangle which has
twosides parallel to the coordinate axes.

(4) Now I shall prove the theorem for any triangle.
Let $T$ be a triangle. Any shape of triangle can be enclosed
in a rectangle with edges parallel to the coordinate axes, as in
the diagram, by adding at most three triangles.
From part (1): $$ F(T+T1+T2+T3) = F(T) + F(T1) + F(T2) +
F(T3). $$ But from part (2): $$ F(T+T1+T2+T3) = 0 $$ and from part
(3): $$ F(T1) = F(T2) = F(T3) = 0. $$ From these relations, I
observe that $F(T) = 0$.

(5) As any polygon can be split into triangles and rectangles, the
theorem is proved.