Find an example about the connectedness. Let $I=[0,1]$ and $I\times I$ be a subspace of $\Bbb R^2$.
Find an example satisfying the following conditions:


*

*$A$ and $B$ are subsets of $I\times I$.  

*$A$ and $B$ are connected.  

*$(0,0), (1,1)\in A$ and $(0,1), (1,0)\in B$.  

*$A \cap B=\varnothing$.

 A: The standard example of a set which is connected but not path connected is $(x, \sin(1/x))$ for $0 < x < 1$ together with a second piece such as $(0,y), -1 \le y \le 1.$  You can use this to solve the problem.  Set $A$ to be the union of two sine curves $\sin(1/x)$ rotated $90$ degrees together with one extra point. In more detail, $(x, \sin(1/x))$ with $0 < x \le 1$ has a "tail" at $(1, \sin(1))$ and approaches a "wall" at $(0, y), -1 \le y \le y$.   Rotate this curve so its tail is at $(0,0)$ and its wall is the the line from $(1,0)$ to $(0,1).$   Make a second copy whose tail is at $(1,1)$ and whose wall is again the line from $(1,0)$ to $(0,1)$, but coming from the other side. Their union is not connected without adding a piece of their common wall, so add in the point $(1/2, 1/2)$.  The union of those three pieces is connected.   Take $B$ to be everything else:  $B = I \times I - A$.
To show that $B$ is connected, let $L$ be the wall, the line from $(0,1)$ to $(1,0)$.  Points in $I\times I - L$ that are below and to the right (BR) of the sine curves can be connected with a path.  And similarly for points above and to the left (UL).  So if there is a disconnection of $B$ as $ U \cup V$, then one of them (say $U$) contains BR, and $V$ contains UL.  What about a point $p \in L - (1/2,1/2)$?  It must be in either $U$ or $V$, but every neighborhood of $p$ contains points of BR and UL,and so contains points of $U$ and $V$, contradicting $U \cap V = \emptyset$
A: David Goldberg was definitely along the right track. We can complete the idea as follows. (Edit: the first answer I gave was written in intuitive language, with the idea that the formal mathematical details would be routine to fill in. After reading a comment below, I decided to completely rewrite the answer, so as to give a more formal, precise description.) 
Define a connected set $P = P_1 \cup P_2 \cup P_3 \cup P_4 \cup P_5$ where   
$$P_1 = \{(x, \sin(1/x): 0 < x \leq 1\}$$ 
$$P_2 = \{(t + 2(1-t), t\sin(1)): 0 \leq t \leq 1\}$$ 
$$P_3 = \{(0, 0)\}$$ 
$$P_4 = \{(x, \sin(1/x): -1 \leq x < 0\}$$ 
$$P_5 = \{(-t - 2(1-t), t\sin(-1)): 0 \leq t \leq 1\}.$$ 
Here $P_1$ and $P_4$ are right-hand and left-hand branches of a topologists' sine curve, $P_2$ and $P_5$ are line segments which connect these branches to the points $(2, 0)$ and $(-2, 0)$ respectively, and $P_3$ is just some accumulation point along the $y$-axis that we adjoin so that $P$ is connected. The choice $P_3 = \{(0, 0)\}$ is thus somewhat arbitrary. 
Somewhat similarly, define a connected set $Q = Q_1 \cup Q_2 \cup Q_3 \cup Q_4 \cup Q_5$ where  
$$Q_1 = \{(x, \sin(1/x) + 1/4): 0 < x \leq 1\}$$ 
$$Q_2 = \{(t, t(\sin(1)+1/4) + 2(1-t)): 0 \leq t \leq 1\}$$  
$$Q_3 = \{(0, 1/2)\}$$ 
$$Q_4 = \{(x, \sin(1/x) - 1/4): -1 \leq x < 0\}$$ 
$$Q_5 = \{(-t, t(\sin(-1)-1/4) - 2(1-t)): 0 \leq t \leq 1\}$$ 
Here $Q_1$ is a right-hand branch of the topologists' sine curve but translated up by $1/4$, which has accumulation points $(0, t)$ with $-3/4 \leq t \leq 5/4$. 
Similarly, $Q_4$ is a left-hand branch translated down by $1/4$, which has accumulation points $(0, t)$ with $-5/4 \leq t \leq 3/4$. Clearly the point of $Q_3$ is an accumulation point of each branch. Finally, $Q_2$ and $Q_5$ are line segments which connect these branches to the points $(0, 2)$ and $(0, -2)$, respectively. The set $Q$ is connected for the same reason $P$ is connected. 
Readers may wish to draw a picture. It is clear by construction that $P$ and $Q$ do not intersect (particularly, the topologist-sine-curve branches do not intersect because they are vertical translates of each other). 
Finally, we apply a homeomorphism which will map the four diamond points $(2, 0)$, $(-2, 0)$, $(0, 2)$ and $(0, -2)$ to the four corners of $I \times I$, and map $P \cup Q$ into $I \times I$. For example, we may first apply a rotation which sends the diamond points to the four points $(\pm \sqrt{2}, \pm \sqrt{2})$, followed by scalar multiplication by $1/2\sqrt{2}$ to map to the four points $(\pm 1/2, \pm 1/2)$, followed by a translation $(x, y) \mapsto (x + 1/2, y + 1/2)$. 
