Construct $2$ connected sets $P,Q$ in square abcd, such that $a,c\in P$ and $b,d\in Q$ and $P\cap Q=\emptyset$ Construct $2$ connected sets $P,Q$ in square abcd, such that $a,c\in P$ and $b,d\in Q$ and $P\cap Q=\emptyset$
I claim that the set $P$ containing the line connecting a to b and the line connecting b to c, and the set $Q$ contains the line connecting c to d and the line connecting d to a. Is this correct?
If so, then what is left to do is show that these sets are connected. First, I would split the set $P$ to be $P_1$ which contains the line connecting a to b, $P_2$ which contains the line connecting b to c. Obviously, $P_1$ and $P_2$ are connected. Hence $P$ is connected. The same argument can be applied to $Q$. Hence $Q$ is also connected. Is this correct?
 A: First note that $P$ and $Q$ cannot both be path-connected: this would mean $P$ contains an $a-c$ path, and $Q$ contains a $b-d$ path, which would have to intersect at some point, by continuity. This motivates us to consider something like the topologist's comb: let $a=(0,1),$ $b=(1,1),$ $c=(1,0)$, and $d=(0,0)$ (the corners of the unit square), and let $K=\{1/n:n\in\mathbb{Z},n\geq 2\}.$ Then if we let $P=((0,1]\times\{0\})\cup(K\times [0,1))\cup\{a\}$, and $Q=[0,1]^{2}\setminus P,$ we obtain the desired sets. 
A: More generally (but very non-constructively):
Theorem. Let $(X,\tau)$ be an infinite topological space of cardinality  $|X|=\kappa$ satisfying the following conditions:
(1) $\ |\tau|=\kappa;$
(2) $\ \emptyset\ne U\in\tau\implies|U|=\kappa;$
(3) $\ S\subseteq X,\ |S|\lt\kappa\implies X\setminus S$ is connected.
Then, for any sets $A,B\subseteq X$ with $A\cap B=\emptyset$ and $|A\cup B|\lt\kappa,$ there exist connected sets $P,Q\subseteq X$ with $A\subseteq P,\ B\subseteq Q,$ and $P\cap Q=\emptyset.$
Remark. All the assumptions hold if $X$ is the square $[0,1]\times[0,1]$ with the usual topology, $\kappa=2^{\aleph_0},\ A=\{(0,0),(1,1)\},\ B=\{(0,1),(1,0)\}.$
Proof. Let $\{(U_\alpha,V_\alpha):\alpha\lt\kappa\}$ be the set of all ordered pairs $(U,V)$ of nonempty open sets. We define $p_\alpha,q_\alpha\in X\ (\alpha\lt\kappa)$ by transfinite induction, as follows. Consider an ordinal $\alpha\lt\kappa,$ and suppose $p_\beta,q_\beta$ have been defined for all $\beta\lt\alpha.$ Let $Y_\alpha=A\cup B\cup\{p_\beta:\beta\lt\alpha\}\cup\{q_\beta:\beta\lt\alpha\}$ and let $W_\alpha=(U_\alpha\cap V_\alpha)\cup[X\setminus(U_\alpha\cup V_\alpha)].$ Then $|W_\alpha|=\kappa$ because, if $U_\alpha\cap V_\alpha\ne\emptyset,$ then $|U_\alpha\cap V_\alpha|=\kappa;$ on the other hand, if $U_\alpha\cap V_\alpha=\emptyset,$ then $U_\alpha\cup V_\alpha$ is disconnected, so $|X\setminus(U_\alpha\cup V_\alpha)|=\kappa.$ Since $|Y_\alpha|\lt\kappa,$ it follows that $|W_\alpha\setminus Y_\alpha|=\kappa.$ Therefore we can choose $p_\alpha,q_\alpha\in W_\alpha\setminus Y_\alpha,\ p_\alpha\ne q_\alpha.$ Let $P=A\cup\{p_\alpha:\alpha\lt\kappa\},\ Q=B\cup\{q_\alpha:\alpha\lt\kappa\}.$
It is clear from the construction that $A\subseteq P,\ B\subseteq Q,$ and $P\cap Q=\emptyset;$ we have to show that $P$ and $Q$ are connected. Assume for a contradiction that one of them, say $P,$ is disconnected. Then there are open sets $U,V$ such that $P\subseteq U\cup V,\ P\cap U\cap V=\emptyset,\ P\cap U\ne\emptyset,\ P\cap V\ne\emptyset.$ Now we have $U=U_\alpha,\ V=V_\alpha$ for some $\alpha\lt\kappa,$ and $p_\alpha\in W_\alpha=(U\cap V)\cup[X\setminus(U\cup V)].$ But $p_\alpha\in X\setminus(U\cup V)$ is impossible because $p_\alpha\in P\subseteq U\cup V,$ and $p_\alpha\in U\cap V$ is impossible because $P\cap U\cap V=\emptyset.$ We have arrived at a contradiction.
P.S. In the same way, given pairwise disjoint sets $A_i\ (i\in I)$ with $|I|\lt\kappa$ and $|\bigcup_{i\in I}A_i|\lt\kappa,$ we can construct pairwise disjoint connected sets $P_i\ (i\in I)$ with $A_i\subseteq P_i.$
