A product of connected spaces is connected in the product topology, and that argument is a theorem of ZF; for an infinite index set it is the assertion that the product of nonempty spaces is nonempty that uses the Axiom of Choice

If $P = \varnothing$ then $P$ is connected, no separation of the empty space existing, and the theorem holds; so assume $P \ne \varnothing$ and fix a single point $a \in P$.

For a function $\sigma$ with domain a natural number $n$ and values in $I$, put $P_\sigma := \{\, x \in P : x_j = a_j \text{ for every } j \in I \setminus \sigma[n] \,\}$, the points agreeing with $a$ outside the finite set $\sigma[n]$.

For $x \in P$ and $i \in I$ let $T_{x,i} := \{\, y \in P : y_j = x_j \text{ for every } j \ne i \,\}$, the $i$-th axis through $x$; the map $t_{x,i} : X_i \to P$ sending $u$ to the point with $i$-th coordinate $u$ and $j$-th coordinate $x_j$ for $j \ne i$ has image $T_{x,i}$.

Each $t_{x,i}$ is continuous, since its $i$-th component is the identity of $X_i$ and its $j$-th component for $j \ne i$ is constant, so [A2] applies; hence $T_{x,i}$ is a connected subset of $P$ by [A3], $X_i$ being connected.

By induction on $n \in \mathbb{N}$ using [A6]: for every function $\sigma$ with domain $n$ and values in $I$, the set $P_\sigma$ is connected. At $n = 0$ the domain is empty, $\sigma[0] = \varnothing$, and $P_\sigma = \{a\}$, a singleton, connected by [A4].

For the step, let $\sigma$ have domain $n+1$, let $\tau := \sigma|_n$ and let $i := \sigma(n)$; then $\sigma[n+1] = \tau[n] \cup \{i\}$, so $P_\tau \subseteq P_\sigma$ and every $T_{x,i}$ with $x \in P_\tau$ is contained in $P_\sigma$, since a point of it agrees with $x$, hence with $a$, off $\tau[n] \cup \{i\}$.

Moreover $P_\sigma = P_\tau \cup \bigcup_{x \in P_\tau} T_{x,i}$: given $y \in P_\sigma$, the point $x$ obtained from $y$ by resetting the $i$-th coordinate to $a_i$ lies in $P_\tau$, and $y \in T_{x,i}$.

So, assuming inductively that $P_\tau$ is connected, each $T_{x,i}$ with $x \in P_\tau$ is connected by step 2.1 and meets $P_\tau$ in $x$, whence $P_\sigma$ is connected by [A4] and step 2.4; by [A6] this proves the claim of step 2.2 for every $n$.

Let $D := \bigcup \{\, P_\sigma : \sigma \text{ a function from a natural number into } I \,\}$, the set of points of $P$ agreeing with $a$ outside a finite subset of $I$; every $P_\sigma$ is connected by step 3.1 and contains $a$, so $D$ is connected by [A4].

$D$ is dense in $P$: let $B = \prod_i U_i$ be a nonempty basic open set as in [A1], with $U_i = X_i$ off a finite $F \subseteq I$, and fix $y \in B$; write $F = \sigma[n]$ for a bijection $\sigma : n \to F$, which exists by [A6]; the point $x$ with $x_j := y_j$ for $j \in F$ and $x_j := a_j$ otherwise lies in $P_\sigma \subseteq D$, and lies in $B$, since $x_j = y_j \in U_j$ for $j \in F$ and $x_j \in X_j = U_j$ otherwise.

Hence $\overline{D} = P$ by [A5], and $D \subseteq P \subseteq \overline{D}$, so $P$ is connected by [A5] and step 4.1.