Let the side of the square be $ a$. Then let the sides of one of the parts(the part that is beetwen the other two) be $ x,y$. We know the following $ xy=1$ $ (a-y)x=ax-xy=1$ $ (a-x)y=ay-xy=1$ If we look at the last two (and put in the first one) $ ax-1=1$ $ ay-1=1$ It follows that $ x=y$. And from here $ x=y=1$ $ a-1=1$ so $ a=2$ so the area is indeed $ 4$.

@Wichking. the perpendicular lines are not necessarily parallel to sides of square.

Hmmm... Name the points as in the figure above. O is the center of the square. GH // NM and IJ // PQ. Y and Z are the lines' intercepts. WLOG, assume that E lies in the left section of the square (in the pentagon OGAI). Denote [polygon] be the area of the polygon. It's easy to see that [MEPB] <= [HYPB] <= [MEQC]. Equality holds when E lies on GH. Do this similarly to the remain parts. Then, we can show that there are 3 parts of the square having equal areas iif E is the center, which means the area of the square is 4.

Is my solution correct?

That solution seem to lose some generality, supose that Q lie on the segment [AD]. Then it's to hard to prove the inequality [EYHM]=<[YHCQ]. It's an god solution, anyway. Those geometric problemas are very hard, it seems that is no way to prove...

We name the lines a and b. a and b are perpendicular, and each divides the square in two parts, one of which has area 2. This means that line b is a 90º rotation of a about the center of the square, O. We draw the other 90º rotation, c, and the 180º rotation, d. This divides the square in 9 parts. Because each line is a rotation about O of the other lines, there are two groups of four parts that are rotations of the other ones, and therefore have the same area. Let the area of those parts be r and s, and the area of the central part be t. a and b divide the square in four parts of area r, r+s, r+s and r+2s+t. Three of them can only be equal if s=0, which means that b and c are the same line, and so are a and d, and therefore the central part has area 0. All four parts have area r=1.

Label $P_1P_2P_3P_4$ the given square and $M$ is a point inside it. A line through $M$ cuts $\overline{P_1P_2},\overline{P_3P_4}$ at $P,T$ and another line through $M$ perpendicular to $PT$ cuts $\overline{P_2P_3},\overline{P_4P_1}$ at $S,U.$ $O$ is the center of $P_1P_2P_3P_4.$ Parallel through $O$ to $PT$ cuts $\overline{P_1P_2},\overline{P_3P_4}$ at $E,G$ and parallel through $O$ to $SU$ cuts $\overline{P_2P_3},\overline{P_4P_1}$ at $F,H.$ Let $X \equiv EG \cap SU,$ $Y \equiv FH \cap TP.$ Denote $S,$ $S_A,$ $S_B,$ $S_C,$ $S_D,$ $S_R$ the areas of $P_1P_2P_3P_4,$ $UHYM,$ $PEOY,$ $SFOX,$ $TGXM,$ $OXMY,$ respectively. WLOG assume that $O$ lies inside the quadrilateral $MPP_2S.$ Clearly, the lines $EG$ and $FH$ divide $P_1P_2P_3P_4$ into four quadrilaterals with equal areas. Therefore, we have $\frac{_1}{^4}S-S_B+S_A=\frac{_1}{^4}S-S_A-S_C-S_R=\frac{_1}{^4}S-S_D+S_C=1.$ From these expressions we obtain $S_A+S_D=S_B+S_C$ $\Longrightarrow$ parallelograms $PEGT$ y $SFHU$ have equal areas. But since they have equal altitudes, then it follows that $PE=FS$ $\Longrightarrow$ $OXMY$ is a square. Thus, $S_A=S_C$ and $S_B=S_D.$ Again, using the expressions found previously, we obtain $S_A=S_B=S_C=S_D=0$ $\Longrightarrow$ $O \equiv M$ and the conclusion follows.