Product topology = topology with product metric 
Given metrics $d$ and $e$ on sets $X$ and $Y$ , let $f$ be the product metric on $X \times Y$.
  Prove that the product topology $T_d \times T_e$ on $X \times Y$ is the same as $T_f$.

A product metric $f$, on $X \times Y$ is defined as $$f( (x_1, y_1), (x_2, y_2) ) = Max\{d(x_1, y_1), e(x_2, y_2)\}$$ for $(x_i, y_i) \epsilon X \times Y$
I keep running into circular reasoning when doing this
 A: Hint: The product topology in $X\times Y$ has as basis $U\times V$ where $U$ is open in $X$ and $V$ is open in $Y$ while the metric topology in $X\times Y$ is has as basis the open balls $B_f((x,y),\epsilon)$ where $(x,y)\in X\times Y$ and $\epsilon>0$. You have to prove that both basis give rise to the same topology. There is a theorem to do this:
If $\mathbb{B}$ and $\mathbb{B}'$ are basis for $X$ then they give rise to the same topology on $X$ iff for all $x$ in $X$ and for all $B\in \mathbb{B}$ with $x\in B$ there is $B'\in \mathbb{B}'$ with $B\subseteq B'$ and for all $x$ in $X$ and for all $B'\in \mathbb{B}'$ with $x\in B'$ there is $B\in \mathbb{B}$ with $B'\subseteq B$.
A: How did you define product topology? 
Anyway, you should know that a set $Q$ in the product topology of the product is open (in case of finitely many factors, but you have only two)  if and only if  to every $(x,y)\in Q$ there are open neighbourhoods $U_x, V_y$ of $x, y$ respectively such that $ U_x\times V_y \subset Q$. Since both factors are metric spaces this is true if and only if to each $(x,y) \in Q$ there are real numbers $r,s$ such that the product of the open balls is contained in $Q$, that is $B_r(x) \times B_s(y)\subset Q$. This implies that that the ball of radius $\min(r,s)$ is a subset of $Q$ with respect to the product metric.
There is only a little step missing, the other direction, when you start out with a set which open with respect to the product metric. You should finish this yourself.
