First Solution:
If a polygon does not intersect itself, then by moving clockwise (or counterclockwise) along its sides, the inside (or outside) of the polygon must always be at our right-side. But if the said polygon doesn't intersect itself, then the right-side of its rightmost side (because it is drawn upward) is its outside, and the right-side of its leftmost side (because it is drawn upward) is its inside, which is a contradiction.

Second Solution:
Suppose that the leftmost side of the polygon connects the vertex $A$ upward to the vertex $B$, and the rightmost side connects the vertex $C$ upward to the vertex $D$. Then all of the polygon must be between the lines $AB$ and $CD$. Clearly, the remaining of the polygon contains a path that connects $B$ to $C$, and another one that connects $D$ to $A$; but these two paths intersect.

Third Solution:
Suppose that this polygon has $2n$ sides and doesn't intersect itself. In this case, total sum of the angles of this polygon is equal to $(2n - 2) \times 180^{\circ} = (n - 1) \times 360^{\circ}$. On the other hand, each non-vertical side of this polygon is adjacent to two vertical sides that are drawn upward. So the sum of the two adjacent angles to each non-vertical side is $360^{\circ}$, and since we have $n$ non-vertical sides, the total sum of all angles equals $n \times 360^{\circ}$, which is a contradiction.