Distance between closed and compact sets.

This question is (1-21)(b) from M. Spivak’s Calculus on Manifolds.

Question:
If $A$ is closed, $B$ is compact, and $A \cap B = \emptyset$, prove that there is $d > 0$ such that $||y – x|| \geq d$ for all $y \in A$ and $x \in B$.

Now, I interpret this as an instruction to find a single $d$ that works for all $y \in A$ and $x \in B$. However, I can’t see why the following is not a counter-example:

Consider the set
$$A_0 = (-\infty, 0) \cup \left[\bigcup_{n=1}^{\infty} \left(\frac{1}{n + 1}, \frac{1}{n}\right)\right] \cup (1, \infty)$$
where $(a,b)$ denotes the open interval as usual. Since $A_0$ is a union of open sets, it too is open. Thus
$$A = \mathbb{R} – A_0 = \left\{ \frac{1}{n} \quad \colon \quad n \in \mathbb{N}\right\}$$
is closed. The set
$$B = [-1, 0]$$
is certainly compact. Moreover, $A \cap B = \emptyset$. However, for all $d > 0$, there exists a $y \in A$ such that
$$||0 – y|| = ||y|| < d$$

I must be overlooking something somewhere. Any help spotting where will be appreciated.

Solutions Collecting From Web of "Distance between closed and compact sets."

Here we prove the result of the book:

Recall that the function $x\mapsto d(x,A)$ is continuous and that (since $A$ is closed):
$$x\in A\iff d(x,A)=0$$

$$d=\inf_{x\in B}d(x,A)$$

The function

$$f:B\to \mathbb{R}\quad,\quad x\mapsto d(x,A)$$
is continuous on the compact $B$ then it’s bounded and there’s $x_0\in B$ s.t
$$f(x_0)=\min_{x\in B}f(x)=d=d(x_0,A)>0$$
since $x_0\not\in A$

The counter-example fails as the set $A$ contains $0$ so $A \cap B \ne \emptyset$. I had overlooked this fact for some reason.