What does it mean that two different metrics may define the same collection of open sets?
The assumption is that a given set is equipped with two different metrics to form two different metric spaces.
Does it simply mean that an open subset of a set is independent of a metric?
If so, why??? if not, why???
On
If two metrics are bounded by positive constant multiples of each other, i.e. $cd_1(c,y) \leq d_2(x,y) \leq Cd_1(x,y)$ then they will generate the same open sets. To see that not all metrics generate the same open sets, consider the standard metric on ${\mathbb R}$ compared to the metric that assigns distance 1 between any two distinct numbers.
On
just to look at a slightly less detailed level
one way of seeing a topology (on a set $X$) is in terms of a basis of open sets. the open sets defined by a basis $B$ are arbitrary unions of sets belonging to the basis.
suppose you have two topologies for $X$ with bases $B_1$ and $B_2$. then the topologies are equivalent if and only if the bases interpenetrate - in the sense that for any $b_1 \in B_1$ we can find a $b_2 \in B_2$ such that $b_2 \subset b_1$ and vice versa.
in metric spaces we define the topology in terms of a basis of open balls of arbitrary small radius. so given a $b_1$ we may join with Hamlet's dyslexic brother in asking "$b_2$ or not $b_2$? that is the question!" (and vice versa, of course)
Let $X$ be a set and $d_1$ and $d_2$ metrics on $X$. That $d_1$ and $d_2$ generate the same collection of open sets means that each set $A\subset X$ is open with respect to $d_1$ if and only if it is open with respect to $d_2$. One example of this case is given in a comment. Conversely, consider $X=\mathbb R$ and $d_1(x,y):=|x-y|$ and $$ d_2(x,y):=\begin{cases}0,&x=y \\ 1,&x\neq y\end{cases}. $$ One can show that each set $A\subset\mathbb R$ is open with respect to $d_2$ (try to prove this as an exercise), but $A=\{0\}$ for example is not open with respect to $d_1$.