[proofplan] We lift a separation of $X$ to a separation of the product $X \times Y$. Since $Y$ is non-empty, each non-empty piece of the separation of $X$ gives a non-empty product piece. The definition of the product topology makes these product pieces open, and the set-theoretic identities for products show that they are disjoint and cover $X \times Y$. [/proofplan] [step:Choose a separation of $X$] Since $(X,\tau_X)$ is disconnected, by the definition of disconnectedness there exist subsets $U,V \subset X$ such that $U \neq \varnothing$, $V \neq \varnothing$, $U \cap V = \varnothing$, $U \in \tau_X$, $V \in \tau_X$, and $U \cup V = X$. [/step] [step:Lift the separation to two non-empty open subsets of $X \times Y$] Define subsets $A,B \subset X \times Y$ by $A = U \times Y$ and $B = V \times Y$. Because $U \neq \varnothing$ and $Y \neq \varnothing$, choose $u_0 \in U$ and $y_0 \in Y$. Then $(u_0,y_0) \in A$, so $A \neq \varnothing$. Because $V \neq \varnothing$ and $Y \neq \varnothing$, choose $v_0 \in V$ and $y_1 \in Y$. Then $(v_0,y_1) \in B$, so $B \neq \varnothing$. Since $U \in \tau_X$ and $Y \in \tau_Y$, the rectangle $U \times Y$ is open in the [product topology](/page/Product%20Topology) $\tau_X \times \tau_Y$. Hence $A$ is open in $X \times Y$. Since $V \in \tau_X$ and $Y \in \tau_Y$, the rectangle $V \times Y$ is open in the product topology $\tau_X \times \tau_Y$. Hence $B$ is open in $X \times Y$. [guided] We want to turn the separation of $X$ into a separation of $X \times Y$. The natural candidates are the full vertical slabs over the two separating pieces of $X$. Define subsets $A,B \subset X \times Y$ by $A = U \times Y$ and $B = V \times Y$. First we verify non-emptiness. Since $U \neq \varnothing$, there exists $u_0 \in U$. Since $Y \neq \varnothing$, there exists $y_0 \in Y$. Therefore $(u_0,y_0) \in U \times Y = A$, so $A \neq \varnothing$. Similarly, since $V \neq \varnothing$ and $Y \neq \varnothing$, there exist $v_0 \in V$ and $y_1 \in Y$. Therefore $(v_0,y_1) \in V \times Y = B$, so $B \neq \varnothing$. Now we verify openness in the product topology. The [product topology](/page/Product%20Topology) $\tau_X \times \tau_Y$ is generated by rectangles $O_X \times O_Y$ with $O_X \in \tau_X$ and $O_Y \in \tau_Y$. Because $U \in \tau_X$ and every topology contains the whole space, $Y \in \tau_Y$. Hence $U \times Y$ is one of the basic open rectangles in $X \times Y$, so $A$ is open. The same argument applies to $V$: since $V \in \tau_X$ and $Y \in \tau_Y$, the set $V \times Y = B$ is open in $X \times Y$. [/guided] [/step] [step:Verify that the lifted sets are disjoint and cover the product] We compute their intersection using the definition of Cartesian product: $A \cap B = (U \times Y) \cap (V \times Y) = (U \cap V) \times Y = \varnothing \times Y = \varnothing$. Thus $A$ and $B$ are disjoint. We compute their union using distributivity of Cartesian product over union: $A \cup B = (U \times Y) \cup (V \times Y) = (U \cup V) \times Y = X \times Y$. Thus $A \cup B = X \times Y$. [/step] [step:Conclude that $X \times Y$ is disconnected] The subsets $A$ and $B$ are non-empty, open in the product topology $\tau_X \times \tau_Y$, disjoint, and satisfy $A \cup B = X \times Y$. Therefore they form a separation of $X \times Y$. By the definition of disconnectedness, $X \times Y$ is disconnected in the product topology. [/step]