The category of graphs and homomorphisms has countable powers. Given a graph $H = (V(H), E(H))$, its countable power is the graph $H^\omega = (V(H^\omega), E(V^\omega))$ where $V(H^\omega) = V(H)^\omega$ is the countable power of vertices, and $E(H^\omega)$ is defined for $a, b \in V(H)^\omega$ by
$$\{a, b\} \in E(H^\omega) \iff
\forall i \in \omega \,.\, \{a_i, b_i\} \in E(H).
$$
For each $i \in \omega$ there is a projection $\pi_i : H^\omega \to H$, defined by $\pi_i(a) = a_i$.
With these projections we really get a power:
$\mathrm{Hom}(L, H^\omega) \cong \mathrm{Hom}(L, H)^\omega$
holds in virtue of the isomorphism taking $f : L \to H^\omega$ to $(\pi_i \circ f)_{i \in \omega}$.
For any graph $H$ we have $H^\omega \cong H^\omega \times H^\omega$, with isomorphism taking $a \in V(H^\omega)$ to $((a_{2 i})_{i \in \omega}, (a_{2 j + 1})_{j \in \omega})$.
Now consider the graph $G = K_3^\omega$ where $K_3$ is the complete graph on three vertices $V(K_3) = \{0, 1, 2\}$. Given $a, b \in \{0,1,2\}^\omega$, we have
$$\{a, b\} \in E(K_3^\omega) \iff \forall i \in \omega \,.\, a_i \neq b_i.$$
The only question remaining is whether $G$ is connected. Take arbitrary vertices $a, b \in \{0,1,2\}^\omega$ and let $c \in \{0,1,2\}^\omega$
be defined by $c_i = \min (\{0, 1, 2\} \setminus \{a_i, b_i\})$. Then $c_i \neq a_i$ and $c_i \neq b_i$, for all $i \in \omega$, therefore $a$ and $b$ are connected via $c$. The graph $G$ is connected.