Finite Product Space is Connected iff Factors are Connected/Basis for the Induction

Theorem
Let $T_1$ and $T_2$ be topological spaces.

Then the product space $T_1 \times T_2$ is connected iff $T_1$ and $T_2$ are connected.

Necessary Condition
Let $T_1 \times T_2$ be connected.

By Projection from Product Topology is Continuous, $T_1$ and $T_2$ are continuous images under the projections $\operatorname{pr}_1$ and $\operatorname{pr}_2$.

Hence by Continuous Image of Connected Space is Connected, $T_1$ and $T_2$ are connected.

Sufficient Condition
Suppose that $T_1$ and $T_2$ are connected.

Let:
 * $I = T_2$
 * $\forall y \in T_2: C_y = T_1 \times \left\{{y}\right\}$
 * $B = \left\{{x_0}\right\} \times T_2$ for some fixed $x_0 \in T_1$.

Each $C_y$ is homeomorphic to $T_1$ by Topological Product with Singleton.

By Connectedness is a Topological Property, each $C_y$ is therefore connected.

By the same argument, $B$ is also connected.

Also:
 * $C_y \cap B = \left\{{\left({x_0, y}\right)}\right\}$ and hence is non-empty
 * $\displaystyle T_1 \times T_2 = B \cup \bigcup_{y \mathop \in T_2} C_y$.

So by the corollary to corollary to Space with Connected Intersection has Connected Union, it follows that $T_1 \times T_2$ is connected.

Also see

 * Product Space is Path-connected iff Factor Spaces are Path-connected