Say that I have the set $A=\cup_{n=1}^{\infty}[n,n+\frac{1}{n}]$. How do I show that this is a Borel set?
My attempt: I know that $\sigma(\mathscr{L})=\mathscr{B}(\mathbb{R})$ where $\mathscr{L}$ is the family of closed sets. Since $\sigma(\mathscr{L})$ has a family $(A_n)_{n\in\mathbb{N}}$ where $A_n:=[n,n+\frac{1}{n}]$, it follows from the definition of a $\sigma$-algebra that $\cup_{n=1}^{\infty}A_n=\cup_{n=1}^{\infty}[n,n+\frac{1}{n}]\in\sigma(\mathscr{L})=\mathscr{B}(\mathbb{R})$. Members of the Borel $\sigma$-algebra are Borel sets, hence $A$ is a Borel set.
Your solution is good. Actually what you have proved is that the $F_\delta$ sets are borelian. This follows, as you have showed, nicely from the definition of the smallest $\sigma$-algebra that contains the open sets of $\mathbb R^n$ under the standard topology.