Using only the definition of a compact set, determine if $\Lambda$ is a compact set.
Let $\Lambda = \big[\frac{1}{2},1\big)$ for $n \in \mathbb{N}-\{1,2\}$. Let $\mathscr{F} = \big\{\big(\frac{1}{n},1-\frac{1}{n}\big) \, : \, n\in \mathbb{N}-\{1,2\}\big\}$. Clearly, $\mathscr{F}$ is an open cover of $\Lambda$. However, if $\mathscr{G} = \big\{\big(0,\frac{1}{2} - \frac{1}{n}\big) \, : \, n\in \{3,4,\dotsc,k\}\big\}$, then $\mathscr{G} \subseteq \mathscr{F}$ but
$$\bigcup_{n=3}^k \enspace \Big(0 \,,\,\frac{1}{2} - \frac{1}{n}\Big) = \Big(0,\frac{1}{2}\Big)$$
where it is plain that $\Lambda \not\subseteq \big(0,\frac{1}{2}\big)$. Hence, $\mathscr{G}$ is not a finite subcover of $\Lambda$, so $\Lambda$ is not compact.
Is the above argument valid? Is there a better solution?
You actually have two problems. First, the elements of $\mathscr{G}$ are actually not elements of $\mathscr{F}$. Second, not all finite subsets of $\mathscr{F}$ are of the form of your $\mathscr{G}$ (i.e. the collection of all intervals with $n=3$ up to $n=k$; some $n$ might get skipped).
The second problem is easy to fix: one can argue that any finite subset of $\mathscr{F}$ is contained in $\{ (1/n,1-1/n) : n \in \{ 3,4,\dots,k \} \}$ for some $k \in \mathbb{N}$. (Just take $k$ to be the maximum index of all the intervals in the finite subset. It's a maximum of finitely many numbers so it is finite.)
The first problem is not so easy to fix while keeping the left endpoints at $0$. But you don't need to do that. Instead try to work with my suggestion above directly.