I'm reading the following two passages and am having trouble understanding how their contrapositives are constructed.
Consider the following statements:
If $X$ is a separable metric space, and $S \subset X$ is a subset such that there exists $d_0>0$ with $d(s,s') \geq d_0$ for all distinct $s,s'\in S$, then $S$ is countable.
This will read: $$(\text{X is separable)&(}(\forall s,s'\in S\subset X)(\exists d_0>0)(d(s,s')>d_0))\implies\text{S is countable}$$
Conversely, if $X$ is a metric space containing an uncountable subset $S$ any two of whose points are at least $d_0$ apart, then $X$ is not separable.
I want to take the contrapositive of the logical implication I wrote previously to conclude the second passage. So far:
$$S \subset X\text{ is uncountable} \implies \neg \big( (\text{X is separable)&(}(\forall s,s'\in S)(\exists d_0>0)(d(s,s')>d_0)) \big)$$
Now I'm stuck.
The first statement is just (written as a logical formula) (your interpretation of it is not quite right):
$$(X \text{ separable }) \to \left(\forall S\subseteq X: (\exists d_0 >0: \forall x,y \in S: d(x,y) > d_0) \to (S \text{ at most countable })\right)$$
So its contrapositive is ($p \to q$ is equivalent to $\lnot q \to \lnot p$):
$$\lnot \left(\forall S\subseteq X: (\exists d_0 >0: \forall x,y \in S: d(x,y) > d_0) \to (S \text{ at most countable })\right) \to (X \text{ not separable })$$
where the first negation can be "expanded" (a $\lnot (\forall x: \phi(x))$ is equivalent to $\exists x : \lnot \phi(x)$ and an implication $p \to q$ is false exactly when $p$ is true and $q$ is false) to:
$$\left(\exists S\subseteq X: (\exists d_0 >0: \forall x,y \in S: d(x,y) > d_0) \land (S \text{ uncountable })\right) \to (X \text{ not separable })$$
which exactly says that if we have an uncountable set all $d_0$ apart, $X$ is not separable, as claimed.