First we show that the area of any triangle is at most 0.5. Consider a square surrounding the triangle. Then from that square, we can remove some rectangles and right triangles to get back the original triangle. The areas of the square, rectangles, right triangles are all multiples of 0.5. Thus the area of the original triangle is a multiple of 0.5, so it is at least 0.5. Now back to the original question. We consider a pentagon with the smallest area. This is possible since the area is a multiple of 0.5. We colour the grid with 4 co ours as such: 121212 343434 121212 343434 Then the pentagon will have 2 vertices of the same colour by pigeonhole principle. So the midpoint of the 2 vertices is also a lattice point. If the midpoint is on the side of the pentagon, then we can remove a vertex and replace it with the midpoint. The resulting pentagon is convex and has a smaller area, contradiction. So now the midpoint must be inside the pentagon. We connect the midpoint with the 5 vertices and we obtain 5 triangles each of area at most 0.5. So the overall area of the pentagon is at least 2.5.

I'm going to solve by Pick's theorem. At first we'll prove the following lemma. lemma If there is no lattice point on the side of the pentagon(except for the vertices),there is at least one lattice point inside of the pentagon(not on the side). proof of lemma)By pigeonhole principal,there are two vertices such that these are congruent in mod2.Then their midpoint is lattice point which is inside of the pentagon and isn't vertice.The proof is completed. Now we consider the problem.If there is lattice point on the side,we can get new pentagon by tying lattice point on the side and some vertice.Then the number of lattice points in the pentagon strictly decrease.So after finite operation,we can get pentagon whose sides don't include lattice point(except for the vertice).This pentagon's area is smaller than original pentagon's area.By lemma,there is at least one lattice point inside of this pentagon.So by Pick's theorem,this pentagon's area $S\ge \frac{5}{2}+1-1=\frac{5}{2}$ Hence the proof is completed.$\blacksquare$

How do you know tiling is always possible? Maybe in some step you'll get a polygon with crossed sides, because of the tiling operation