Let $(X,d)$ be a metric space, and $U\subset{X}$ be an open set. Then $U$ may be written as a union of closed sets.
I think I have a proof for this but I'm not quite sure, I feel like there must be something missing to it, or that it is not fully rigorous. I've seen some posts that proved it differently and more complicated, so I feel like mine might be wrong.
My Proof.
Let $V=\{J\subset{U}: $$J$ is closed$\}$. Obviously, if $x\in{V}$ then $x\in{U}$. If $x\in{U}$ then there exists an open ball $B(x,\delta)$ such that $B(x,\delta)\subset{U}$. Take an $0<\varepsilon<\delta$. Then the closed ball $C(x,\varepsilon)\subset{B(x,\delta)}\subset{U}$ is in V, as $x\in{C(x,\varepsilon)}$ then x is in $V$. So $V=U.$ We take $\bigcup_{J\in{V}}{J}=V=U$ and we are done.
I'm not quite sure if the union business I did at the end is correct, but besides that I can't really find a problem with it.