I want to prove that if a topological space $(X, T)$ is separable then the space has the $CCC$ (i.e., there are is no uncountable collection of mutually disjoint open subsets of $X$.)
My attempt of a proof:
We proceed by contrapositive, namely that $X$ does not have the $CCC$ condition. This implies that there exists an uncountable collection of mutually disjoint subsets, $C$. Suppose $X$ is a separable space. Then there exists a dense countable subset of $D$. Since $D$ is dense there is a non empty intersection with all non empty open sets of $X$ so, that means that all of the sets of $C$ has a non empty intersection with $D$. This means that, since there is only a countable amount of elements of $D$ then there are some sets of $C$ that are intersecting more than one element (since there are uncountable sets that are containing only countable amount of elements). This contradicts that those sets are mutual disjoint and hence $X$ has the $CCC$ condition.
Does that proof make sense, if not where are my mistakes? If it does, how can I make it more concise and clear.
You don't need the contrapositive.
You can just directly argue. Since $D$ is a dense set, it meets every non-empty open set. If we have a collection of pairwise disjoint open sets $\{U_i\mid i\in I\}$, then $\{D\cap U_i\mid i\in I\}$ is a pairwise disjoint family of non-empty subsets of $D$.
We can now choose a point from each $D\cap U_i$, which defines an injection from $\{U_i\mid i\in I\}$ into $D$, so $I$ must be countable.
On the style-front, I'd recommend adding line breaks and using $\rm\LaTeX$ where appropriate.