After doing some homework for my Topology course, I am confused about why this reasoning doesn't show that $[0,1]$ is not compact in $\mathbb{R}$. Clearly this isn't true, so where is the flaw?
Answer: The problem is that $[0,1]-K \neq \bigcup_{i \in \mathbb{N}} \left(\dfrac{1}{n+1},\dfrac{1}{n}\right)$ because $0 \in [0, 1] - K$.
Note that $K = \{\frac{1}{n} | n \in \mathbb{N}\}$ isn't compact because it doesn't have a least element. Additionally, note that $K$ is closed in $[0,1]$ as
$$[0,1] - K = \bigcup_{i \in \mathbb{N}} \left(\dfrac{1}{n+1},\dfrac{1}{n}\right).$$
Since $K$, a subspace of $[0,1]$, is closed and not compact, $[0, 1]$ cannot be compact in $\mathbb{R}$.
Note: However, does this reasoning work in $\mathbb{R}_k$? What about in $\mathbb{R}_l$?
$[0,1]\setminus K$ is the the union you wrote. Note that $0 \in [0,1]\setminus K$ but $0$ is not in the union.