The natural map $i \colon \mathcal O_F \to F$ is an injective map of $\mathbb F_p$-vector spaces, hence we can choose a splitting, i.e. an $\mathbb F_p$-linear map $s \colon F \to \mathcal O_F$ such that $s \circ i$ is the identity.

We have $|s(x)| \leq |x|$ since if $x \in \mathcal O_F$ then $|s(x) | = |x|$ and if $x \notin \mathcal O_F$ then $|s(x)| \leq 1 \leq |x|$.

By $\mathbb F_p$-linearity, if $$c = \sum_{i=1}^n c_{1,i} \otimes c_{2,i}$$ then $$ s \otimes s(c) = \sum_{i=1}^n s(c_{1,i}) \otimes s(c_{2,i})$$ and if $c \in \mathcal O_F \otimes \mathcal O_F$ then by definition $$s \otimes s(c)=c.$$

Combining these, we have

$$|c|_{prod}=\inf\left\{\max_{1\le i\le n}\{|c_{1,i}||c_{2,i}| \}\ : \ c=\sum^{n}_{i=1}c_{1,i}\otimes c_{2,i}, \text{where } c_{1,i}, c_{2,i}\in F\right\} \ge \inf\left\{\max_{1\le i\le n}\{|s(c_{1,i})||s(c_{2,i})| \}\ : \ c=\sum^{n}_{i=1}c_{1,i}\otimes c_{2,i}, \text{where } c_{1,i}, c_{2,i}\in F\right\} = \inf\left\{\max_{1\le i\le n}\{|s(c_{1,i})||s(c_{2,i})| \}\ : \ s\otimes s (c)=\sum^{n}_{i=1}s(c_{1,i})\otimes s(c_{2,i}), \text{where } c_{1,i}, c_{2,i}\in F\right\} = \inf\left\{\max_{1\le i\le n}\{|s(c_{1,i})||s(c_{2,i})| \}\ : \ c=\sum^{n}_{i=1}s(c_{1,i})\otimes s(c_{2,i}), \text{where } c_{1,i}, c_{2,i}\in F\right\} \ge \inf\left\{\max_{1\le i\le n}\{|d_{1,i}||d_{2,i}| \}\ : \ c=\sum^{n}_{i=1}d_{1,i}\otimes d_{2,i}, \text{where } d_{1,i}, d_{2,i}\in {\mathcal{O}_{F}}\right\} = |c|'_{prod}$$

which together with the trivial inequality proves the identity.

The natural map $i \colon \mathcal O_F \to F$ is an injective map of $\mathbb F_p$-vector spaces, hence we can choose a splitting, i.e. an $\mathbb F_p$-linear map $s \colon F \to \mathcal O_F$ such that $s \circ i$ is the identity.

We have $|s(x)| \leq |x|$ since if $x \in \mathcal O_F$ then $|s(x) | = |x|$ and if $x \notin \mathcal O_F$ then $|s(x)| \leq 1 \leq |x|$.

By $\mathbb F_p$-linearity, if $$c = \sum_{i=1}^n c_{1,i} \otimes c_{2,i}$$ then $$ s \otimes s(c) = \sum_{i=1}^n s(c_{1,i}) \otimes s(c_{2,i})$$ and if $c \in \mathcal O_F \otimes \mathcal O_F$ then by definition $$s \otimes s(c)=c.$$

Combining these, we have

$$\begin{aligned}|c|_{prod} &=\inf\left\{\max_{1\le i\le n}\{|c_{1,i}||c_{2,i}| \}\ : \ c=\sum^{n}_{i=1}c_{1,i}\otimes c_{2,i}, \text{where } c_{1,i}, c_{2,i}\in F\right\} \\ &\ge \inf\left\{\max_{1\le i\le n}\{|s(c_{1,i})||s(c_{2,i})| \}\ : \ c=\sum^{n}_{i=1}c_{1,i}\otimes c_{2,i}, \text{where } c_{1,i}, c_{2,i}\in F\right\} \\ &= \inf\left\{\max_{1\le i\le n}\{|s(c_{1,i})||s(c_{2,i})| \}\ : \ s\otimes s (c)=\sum^{n}_{i=1}s(c_{1,i})\otimes s(c_{2,i}), \text{where } c_{1,i}, c_{2,i}\in F\right\} \\ &= \inf\left\{\max_{1\le i\le n}\{|s(c_{1,i})||s(c_{2,i})| \}\ : \ c=\sum^{n}_{i=1}s(c_{1,i})\otimes s(c_{2,i}), \text{where } c_{1,i}, c_{2,i}\in F\right\} \\ &\ge \inf\left\{\max_{1\le i\le n}\{|d_{1,i}||d_{2,i}| \}\ : \ c=\sum^{n}_{i=1}d_{1,i}\otimes d_{2,i}, \text{where } d_{1,i}, d_{2,i}\in {\mathcal{O}_{F}}\right\} \\ &= |c|'_{prod}\end{aligned}$$

which together with the trivial inequality proves the identity.