How do I prove that a filter converges in a cartesian product?

The forward direction is fine, using the continuity of the projections.

For the reverse, we have a point $p = (p_i)_{i \in I}$ from the component spaces and we want to show that $\mathfrak{F}$ converges to $p$. We don't have to use contradiction, but use the standard base of the product topology: a basic open subset $B$ of $p$ is of the form $B=\bigcap_{i \in F} \pi_i^{-1}[O_i]$ where $O_i \subseteq M_i$ is open and contains $p_i$ and $F \subseteq I$ is finite.

Now show that all $\pi_i^{-1}[O_i] \in \mathfrak{F}$ from the fact that the "component filters" $(\pi_i)_\ast(\mathfrak{F})$ converge to $p_i$ for all $i$. Then the filter axioms tell us that $B \in \mathfrak{F}$ and as this holds for all basic open subsets of $p$, we have that $\mathfrak{U}(p) \subseteq \mathfrak{F}$ and so $\mathfrak{F} \to p$.