To verify that given product metric is discrete I would request guidence for checking if my solution to the following problem is correct
Let ($X_{1}, d_{1}$) and ($X_{2},d_{2}$) be two metric spaces. Let X =$X_{1} \times X_{2}$ be the Cartesian product of $X_{1}$  and  $X_{2}$. Then define function d on $X \times X$ by
d(($x_{1},x_{2}$),($y_{1},y_{2}$)) = max{$d_{1}(x_{1},y_{1}$),$d_{2}(x_{2},y_{2}$)}
Verify that the product metric on $X_{1}\times X_{2}$ is disrete.
My attempt is as under:
We need to prove that if 
($x_{1},x_{2}$) = ($y_{1},y_{2}$), then d=$0$ and 
if ($x_{1},x_{2}$) $\neq$  ($y_{1},y_{2}$) then d=$1$
Now if ($x_{1},x_{2}$) = ($y_{1},y_{2}$), then $x_{1}=x_{2}$ and $y_{1}=y_{2}$ which implies
$d_{1}=0=d_{2}$ and therefore d=max{$0,0$}=$0$ 
If ($x_{1},x_{2}$) $\neq$  ($y_{1},y_{2}$) then we have  following three possibilities:
$x_{1}=y_{1}$ but $x_{2}\neq y_{2}$ $\implies$ $d_{1}=0$, $d_{2}=1$ $\implies$ 
d=max{$0,1$}=$1$
$x_{1}\neq y_{1}$ but $x_{2}=  y_{2}$ $\implies$ $d_{1}=1$, $d_{2}=0$ $\implies$ 
d= max{$1,0$}=$1$
$x_{1}\neq y_{1}$ and $x_{2}\neq y_{2}$ $\implies$ $d_{1}=1$, $d_{2}=1$ $\implies$ 
d=max{$1,1$}=$1$
Hence proved.
 A: If you add to you problem that both $d_1$ and $d_2$ are the discrete metric, then, yes, your answer is correct.
A: Almost.  But a discrete metric, as oppose to the discreet metric does not mean that $d (a,b)=1$ if $a\ne b $.
A discrete metric means that for every $x\in X $ then there is a $\delta >0$ (dependant upon $x $) So that for all $y\ne x $ then $d (x,y)>\delta $.
Let $(a,b)\in X_1\times X_2$.  Let $\delta_a >0$ be such that $d_1 (a,x)>\delta_a $ for all $x\ne a $ and let $\delta_b >0$ be such that $d_2 (b,y)>\delta_b $ for all $y\ne b $.
Let $\delta=\min (\delta_a,\delta_b) $.
If $(c,d)\in X_1\times X_2$ and $(c,d)\ne (a,b) $
Then $d ((a,b)(c,d))=  \max (d_1 (a,c),d_2(b,d) )$
Case 1: $a=c ; b\ne d$.  
$ d ((a,b)(c,d))=  \max (0,d_2(b,d)) =d_2 (b,d)>\delta_b \ge \delta $
Case 2: $b=d;a\ne c $
$d ((a,b)(c,d))=  \max (d_1 (a,c),0 )    =d_1 (a,c)>\delta_a\ge \delta$
Case 3: $a\ne c;b\ne d $
$d ((a,b)(c,d))=  \max (d_1 (a,c),d_2(b,d)) > \max (\delta_a,\delta_b)\ge \delta $
So $d $ is a dicrete metric but not the discrete metric.  (Unless $d_1$ and $d_2$ are both the discrete metric.)
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No.  What you have shown is 
IF $d_i$ are the discrete metric on $X_i$ then $d((x,y), (u,v)) = \max (d_1(x,u), d_2(y,v))$ is the discrete metric on $X_1 \times X_2$.
That is not what was asked.
(Added: What was asked is actually false and can not be verified and can be shown to be false.)
Mistake 1:  $d_i$ do not have to be discrete at all.
Mistake 2:  To say $(E,d)$ is discrete does not actually mean that $d$ is "the" discrete metric.
"The" discrete metric is $d(a,b) = 0$ if $a=b$ and $d(a,b) = 1$ if $a\ne b$.
To say a metric space $(E,d)$ is discrete (not the discrete, just discrete) means.
For every $x \in E$ there is $\delta_x > 0$ (which is dependent  on $x$; different points will have different values) so that for all $y \in E$ where $y\ne 0$ then $0  < d(x,y)< \delta_x$.
If a metric space is discrete then for every point $x$ there will be some open neighborhood of $x$ with a radius of $\delta_x > 0$ where $x$ is the only point in the neighborhood.  As a result: Every set is open; every point is isolated; there are no accumulation points.
So to verify the statement:
Assuming we have no idea whether or not $d_1$ or $d_2$ are dicrete or not and even assuming they are not.
We need to verify that for any $(a,b) \in X_1\times X_2$ there is a $\delta > 0$.  so that if $(c,d) \ne (a,b)$ then $d((a,b),(c,d) = \max(d_1(a,c),d_2(b,d)) > \delta$.
Hmmm... 
Since $(a,b) \ne (c,d)$ we have your three conditions
1) $a=c$ and $d((a,b)(c,d)) = d_2(b,d)>0$  but I don't see any reason why this should have any lower bound above $0$.
Frankly.... I don't think this statement is true!
In fact, it can't be.  Simply let $X_1,X_2 = \mathbb R$ and $d_1 = d_2$ be the Euclidean metric.
Then $d_m = \max(d(x_1,x_2), d(y_1,y_2))$.
Let $(a,b) \in \mathbb R^2$ and let $\delta > 0$.  Let $c = a+ \frac {\delta}2$ and $d= b + \frac {\delta}2$.  Then $d(a,c) = \frac {\delta}2 < \delta$ and $d(b,d) = \frac \delta 2$ and $d_m = \max(d(a,c),d(b,c) = \frac \delta 2 < \delta$.
So this is NOT a discrete space.
