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My example is a tad all over the place so any input would be very helpful thank you.

Let A={[0, 1] intersection Q}, for all x belonging to A. Now, let y belong to A and be irrational. Then {(-1, y-1/n): n belonging to N}U(y,2) would be an open cover of A.

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    $[0,1] \cap \Bbb Q$ is unfortunately not closed. Think about $\Bbb Z$ instead. – 2017-02-20
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    Or try $R$ itself. – 2017-02-20
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    Is [0,1] intersect Q compact under R? @Nitrogen – 2017-02-20
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    No, compact subsets of $\Bbb R$ are closed. – 2017-02-20

2 Answers 2

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Consider the set $S = (-\infty, -1] \cup [1, \infty)$. $S$ is a closed subset of $\mathbb{R}$ since $\overline{S} = (-1,1)$, which is an open set.

The family of intervals

$$\{ (n-2, n+2) \mid n \in \mathbb{Z} \}$$

is an open cover of $S$ with no finite subcover.

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Take for example $\mathbb{Z}$ nad the open cover $A=\{(-n,n)|n \in \mathbb{N}\}$

A is an open cover of $\mathbb{Z}$ that does not contain a finite subcover of $\mathbb{Z}$ .

$\mathbb{Z}$ is a closed set with respect to the euclidean topology in the real line.