Compact metrizable space is separable proof question. Prove that a compact metrizable space is separable.  I am confused by a specific case of a compact metrizable space.
Let $[0,1]$ be a compact metrizable space.  Since $[0,1]$ is metrizable, it's Hausdorff. Since $[0,1]$ is compact, $[0,1] \subseteq \bigcup_{x_i \in [0,1]} U_{x_i}$ where each $U_{x_i} \cap U_{x_j} = \emptyset$.  Then $[0,1] \subseteq \bigcup_{n=1}^{N} U_{x_{i_{n}}}$.  But this doesn't make sense because there is no finite collection of pairwise disjoint open sets that covers $[0,1]$.  Could someone explain what I'm misunderstanding here?
 A: The def'n of compactness is not based on pair-wise disjoint open families. A space $S$ is compact iff whenever $F$ is a family of open subsets of $S$ with $\cup F=S,$ there is a finite $G\subset F$ with $\cup G=S.$  It does not matter whether the members of  $F$ or $G$ are pair-wise disjoint or not. (....And if the space $S$ is connected, and $F$ is a cover of $S$ by pair-wise disjoint open sets, then the only possible members of $F$ are $\emptyset$ and $S.$)
Let $S$ be a compact metrizable space and let $d$ be a metric for $S.$ For each $n\in N$ the family $F_n=\{B_d(x,1/n): x\in S\}$ is an open cover of $S.$ 
For each $n\in N $ let $G_n=\{B_d(x,1/n):x\in C_n\}$ be a subcover of $F_n$ (i.e. $G_n\subset F_n$ and $\cup G_n=S$), where $C_n$ is a finite subset of $S.$ It follows that the countable set $T=\cup_{n\in N}C_n$ is dense in $S.$
To prove this, it suffices to show that $T\cap B_d(y,r)\ne \emptyset$ whenever $y\in S$ and $r>0.$ Consider that every $G_n$  is a cover of $S,$ so  take any $n\in N$ with $1/n<r.$  There exists $\sigma\in C_n$ such that $y\in B_d(\sigma,1/n).$ Then $d(y,\sigma)<1/n<r$ so $\sigma \in B_d(y,r).$ Hence $\sigma \in T\cap B_d(y,r).$
