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I have a general question on how to prove a space is not separable.

I read some posts on this site and it seems like it suffices to find an uncountable family of pairwise disjoint open sets to prove a space is not separable. (here: The space of bounded continuous functions are not separable)

Why an uncountable family of pairwise disjoint open sets is enough?

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    Imagine that a dense subset exists, then it intersects every subset on the family you described, so it can't be countable. – 2017-02-09

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A separable space $X$ is a space with a countable dense subset $S$. Label the elements of the $S$ with $x_i$. Since $S$ is dense, $\bar{S}=X$. Thus any open set $U$ must contain some $x_i$. However, if $X$ contains uncountably many disjoint opens, this is not possible, as there must be some $U_\alpha$ not containing any of the $x_i$.

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Let $(U_i)_{i\in I}$be a family of pairwise disjoints open sets, where $I$ is uncountable. If the space is separable , fix $D=\{x_n| n\in \Bbb N\}$ a dense subset.

For each $i\in I$, choose (there is) $n_i\in \Bbb N$, such that $x_{n_i}\in U_i$.

No the map $I\to \Bbb N$, $i\mapsto n_i$ is injective.