Any subset of a metric space is an infinite union of some individual elements of the space? Let $E$ be a metric space such that the set $\{x\}$ is open $ \forall x \in E$. Does the following proposition make sense?  
All subsets of $E$ are open.
Proof: $\forall S \subset E$, there are points $x_i, i \in \mathbb{N}$ such that $S = \cup_{i=1}^{\infty} x_i$. Since the union of open sets is open, any $S$ is open in $E$.

The statement is true, but your argument is false: what if $S$ is uncountable?
The key is that a union of arbitrarily many open sets is still open - can you see a way of writing $S$ as a union of open sets? HINT: each $\{x\}\subseteq S$ is open . . .
