Question: Let $F$ be the interval $(0,1)$ and find an open cover $G$ such that no finite sub-collection of $G$ covers $F$.
I believe I have the answer I would appreciate some reassurance of my answer or if it is wrong a correction.
Answer: Let $F$ be the interval $(0,1)$ and let $G$ be the set $(0,1-\frac{1}{n})$ where $n=1,2,3,\dots$
It is obvious that $G$ covers $F$ but no finite sub-collection of $G$ could because the set $F$ is open.
Let $G_n = (0, 1 - \frac{1}{n})$, then as I pointed out in the comments, $G := \{G_n \mid n\in \mathbb{N}\}$ is a cover for $(0, 1)$ which has no finite subcover (so $(0, 1)$ is not compact).
To see that it is a cover, let $x \in (0, 1)$, then $1 - x > 0$. By the Archimedean property of the real numbers, there is $N \in \mathbb{N}$ such that $\frac{1}{N} < 1 - x$, and hence $x < 1 - \frac{1}{N}$, so $x \in G_N$.
Now consider $\{G_{n_k} \mid k = 1, \dots, K\} \subset G$. Note that as $G_{n_1} \subset G_{n_2} \subset \dots \subset G_{n_K}$, $\bigcup_{k=1}^KG_{n_k} = G_{n_K} = (0, 1 - \frac{1}{n_K}) \subsetneqq (0, 1)$. Therefore $G$ does not have a finite subcover.