Showing that a metric topology on $X \times Y$ is the same as the product $T_{e} \times T_{d}$ 
Question:
Given metric space $\left ( X,d \right )$ and $\left ( Y,e, \right )$,
  define the product metric on $X \times Y$ by $f\left ( \left ( x_{1},y_{1} \right ), \left ( x_{2},y_{2} \right )\right ):=\max\left \{ d\left ( x_{1},x_{2} \right ),e\left ( y_{1},y_{2} \right ) \right \}$
Show that a metric topology $T_{f}$ on $X \times Y$ is the same as the product product $T_{e} \times T_{d}$

Any help is appreciated.
Thanks in advance.
Edit: I think it suffices to show they have the same basis.
 A: You can understand this in a few steps.
First, prove that there is a basis for the product topology on $X \times Y$ consisting of all subsets of $X \times Y$ of the form $B_d(x,r) \times B_e(y,s)$, for $x \in X$, $y \in Y$, $r,s > 0$ (notice that the two radii $r,s$ are allowed to vary independently). This is a generalization of the "box basis" on $\mathbb{R}^2$. It is a special case of a simple but general theorem about the product topology on $X \times Y$: if $\mathcal{B}_X$ is a basis for the topology on $X$ and, and if $\mathcal{B}_Y$ is a basis for the topology on $Y$, then there is a basis $\mathcal{B}$ for the product topology on $X \times Y$ which is the collection of all subsets $B \times B' \subset X \times Y$ such that $B \in \mathcal{B}_X$ and $B' \in \mathcal{B}_Y$. 
Second, use what you know about open balls being a basis for a metric topology. Thus, for each $p = (x_0,y_0) \in X \times Y$ and each $r < 0$ we have a corresponding open ball 
$$B_f(p,r) = \{(x,y) \in X \times Y \quad | \quad |x-x_0| < r \quad\text{and}\quad |y-y_0| < r\}
$$
In other words, $(x,y) \in B_f(p,r)$ $\iff$ $x \in B_d(x,r)$ and $y \in B_e(y,r)$ $\iff$ $(x,y) \in B_d(x,r) \times B_e(y,r)$. The "metric ball basis" for the metric $f$ on $X \times Y$ is therefore all subsets of the form $B_d(x,r) \times B_e(y,r)$ for $x \in X$, $y \in Y$, and $r>0$ (notice that the two radii are the same number $r$).
Third, prove that the two bases $\{B_d(x,r) \times B_e(y,s)\}$ and $\{B_d(x,r) \times B_e(y,r)\}$ generate the same topology on $X \times Y$. For this, you simply have to prove that if $t = \min\{r,s\}$ then
$$B_d(x,t) \times B_e(y,t) \subset B_d(x,r) \times B_e(y,s)
$$
Notice, although it suffices to show that two bases are the same in order to prove that they generate the same topology, that's not necessary. And in this proof, that's not what happened.
A: A useful criterion is as follows: (Munkres, Lemma 2.2) Let $B,B'$ be bases for the topologies $T,T'$ on $X$. Then the following are equivalent:


*

*$T'$ is finer than $T$

*For each $x \in X$ and each basis element $b \in B$ containing $x$, there is a basis element $b' \in B'$ such that $x \in b' \subset b$.



Fix a point $(x,y) \in X \times Y$. I'll use the notation $B^d_\epsilon(x)$ for the open ball of radius $\epsilon$ around $x$ in the space $(X,d)$.
In the product topology $T_d \times T_e$, every basis element containing $(x,y)$ contains a basis element of the form $B^d_\epsilon(x) \times B^e_\epsilon(y)$ for some $\epsilon > 0$. Then $B^f_\epsilon(x,y) \subset B^d_\epsilon(x) \times B^e_\epsilon(y)$ directly from the definition of $f$.
In the reverse direction, any basis element of $T_f$ containing $(x,y)$ contains a basis element of the form $B^f_\epsilon(x,y)$ for some $\epsilon > 0$. And again, $B^d_\epsilon(x) \times B^e_\epsilon(y) \subset B^f_\epsilon(x,y)$ directly from the definition of $f$.
