Prove that $[2,4)$ is not compact. Use the negation of compactness.
Looking to see if this is correct or if there is a more elegant proof.
My solution: To show that $[2,4)$ is not compact, we must show that there exists an open cover, which does not have a finite sub cover. Take the open cover $(2-\frac{1}{n}, 4-\frac{1}{n})$ where $n \in\mathbb{N}$. We must show that this open cover does not have a finite sub cover. $(2-\frac{1}{n}, 4-\frac{1}{n})$ does not have a finite sub cover, because if it did, then there would exist $n_k$ such that $(2-\frac{1}{n_k}, 4-\frac{1}{n_k})$ covers $[2,4)$ for all $n\in\mathbb{N}$. But $(2-\frac{1}{n_k}, 4-\frac{1}{n_k})$ does not cover $(2-\frac{1}{n_{k+1}}, 4-\frac{1}{n_{k+1}})$. Thus $[2,4)$ is not compact.
Your solution is almost right.
Suppose $(2-\frac{1}{n},4-\frac{1}{n})$ has a finite subcover of $[2,4)$ then there exists and $N \in \mathbb{N}$ such that $\cup_{n=1}^{N} (2-\frac{1}{n},4-\frac{1}{n})$ cover $[2,4)$, but that union is $(1,4-\frac{1}{N})$, so points very close to 4 in $[2,4)$ are not in that union.
Indeed, $4-\frac{1}{2N}$ is in $[2,4)$ but not in $(1,4-\frac{1}{N})$.