Understanding a claim -- which interpretation is right? I'm trying to disprove the following claim.

If $A$, as a subspace of $X$, has discrete topology, then $X$ has discrete topology.

The statement If A, as a subspace of X, has discrete topology means all of the subsets of $A$ are open sets in the metric space $(X,d)$ not $(A,d)$, right?
Or does it just mean that $(A,d)$ is a subspace of $(X,d)$ and A has discrete topology --- all subsets of $A$ are open sets in the metric space $(A,d)$?
 A: No, the clause $A$ as a subspace of $X$, means that we consider $(A,d)$ (in the metric case) as a space in its own right, and the fact that this has the discrete topology, means that every point $a$ of $A$ has a ball $B(a,r)$ such that the ball in $A$ is just the singleton set $\{a\}$. So the only point of $A$ that has distance less than $r$ to $a$ is $a$ itself.
E.g. consider $\mathbb{Z}$ in $\mathbb{R}$, usual metric. Then the ball in $\mathbb{Z}$ of radius $1$ around any point of $\mathbb{Z}$ is just that point (although there are many points of the whole space within that distance, there are none other from $\mathbb{Z}$). 
A: There is a counter example for this.
Consider $X=(\mathbb{R},\mathbb{T} )$, where $\mathbb{T}$ is the half-open interval topology in $\mathbb{R}$. Let $Y=X\times X$ with the product topology and $S=\left\{ (x,-x) : x\in X \right\}$ the antidiagonal.
For $(x,y)\in S$, consider the open set in Y given by $[x,+\infty)\times [y,+\infty)=A$. Thus, the intersection $S\cap A=\left\{ (x,y)\right\}$. This means that the product topology in $S$ is the discrete topology; since $Y$ does not have the discrete topology, your claim is false.
