Since no one gave a full answer, here's a proof using the Archimedean Property of $\mathbb{R}$. Thanks for the original comments, and please let me know if anything here is wrong!
To show that $(a, b)$ is not compact, we just need one example of an open cover that has no finite open subcover. We will use the cover $\{ G_n \} = \left\{ \left( a + \frac{1}{n}, b - \frac{1}{n} \right) \mid n \in \mathbb{N} \right\}$ (If this gives us an invalid segment such as $(1, 0)$, treat it an empty element.)
Why does $\bigcup_{n=1}^\infty G_n$ cover $(a, b)$? Well, for any element $x \in (a, b)$, the Archimedean Property provides an $n \in \mathbb{N}$ such that $n > \max \{ \frac{1}{x - a}, \frac{1}{b - x} \}$. Then $nx - na > 1 \text{ and } nb - nx > 1 \Rightarrow na + 1 < nx < nb - 1 \Rightarrow x \in \left( a + \frac{1}{n}, b - \frac{1}{n} \right).$ Thus every element of $(a, b)$ is in $G_n$ for some $n \in \mathbb{N}$, so $(a, b) \subset \bigcup_{n=1}^\infty G_n$.
Why does $\{ G_n \}$ have no finite subcover? Well, for any $i > j$, $a + \frac{1}{i} < a + \frac{1}{j} < b - \frac{1}{j} < b - \frac{1}{i}$, so that $G_i \supset G_j$. Thus for any $k \in \mathbb{N}$ with $k < \infty$, $\bigcup_{n=1}^k G_n = G_k = (a + \frac{1}{k}, b - \frac{1}{k})$. But $(a, b) \not\subset G_k$---since, for example, $a + \frac{1}{2k} \in (a, b)$ but $\notin (a + \frac{1}{k}, b - \frac{1}{k})$.