Proof Verification: Distance Topology on $\mathbb{R}$

Question

Let $A\subset \mathbb{R}$ be a compact set.
Define $$B_j=\left\{x\in\mathbb{R} : \text{dist}(x, A)<\frac{1}{j}, j\in\mathbb{N}\right\}$$ Prove that $$\bigcap_{j=1}^\infty B_j\subset A$$ Proof:
Suppose $$\bigcap_{j=1}^\infty B_j\not\subset A\\\implies \exists x\in\bigcap_{j=1}^\infty B_j\setminus A\\\implies \text{dist}(x,A)>\epsilon\enspace\text{ for some }\epsilon >0$$ By Archimedean property $\exists j_0\in\mathbb{N}$ such that $\epsilon>\frac{1}{j_0}$.$$\implies \text{dist}(x,A)>\frac{1}{j_0}\\\implies x\not\in B_{j_0}$$This is a contradiction to the fact that $x\in\bigcap_{j=1}^\infty B_j$.
Hence, $\bigcap_{j=1}^\infty B_j\subset A$.

Is my proof correct? If not, please provide me the mistake and how to fix it?
Also, If there are any alternative proof?

Answer 1

Direct proof

Let $ x\in \Bbb R$.

$$x\in \bigcap_{n\ge 1}B_n \implies$$

$$(\forall n\ge 1)\; \;\; x\in B_n \implies$$

$$(\forall n\ge 1) \;\;\; 0\le d(x,A)<\frac 1n \implies$$

$$\lim_{n\to+\infty}d(x,A)=0 \implies$$ $$d(x,A)=0 \implies x\in A$$ because $ A $ is closed as pointed by @Brian M. Scott,

thus $$\bigcap_{n\ge 1}B_n\subset A$$