Prove that if $B$ is the set of rationals in $[0,1]$ with a finite subcover, then: $1 \leq \sum_{k=1}^n m^*(I_k)$

Another way to tackle this problem is the following : Let $ \left \{ I_{k} \right \}_{k=1}^{n} $ such that $ \mathbb{Q} \cap [0,1] \subseteq \bigcup^{n}_{k=1} {I_{k}} $

Taking closures implies that $ [0,1] \subseteq \bigcup^{n}_{k=1} \overline{I_{k}} $ Thus $ 1 \leq m(\bigcup^{n}_{k=1} \overline{I_{k}}) \leq \sum_{k=1}^{n}m(\overline{I_{k}})=\sum_{k=1}^{n}m({I_{k}}) $

$ \Longrightarrow \sum_{k=1}^{n}m({I_{k}}) \geq 1 $ which is exactly what we wanted to prove .

There is a mistake in your argumentation.

There are open covers of the rationals in $[0, 1]$ that don't cover the whole interval. For example, let $x_i$ be an enumeration of all those rationals and take the sets of the form $(x_i - 2^{-i - 10}, x_i + 2^{-i - 10})$. The subadditivity of the outer measure shows that these sets don't cover the whole interval $[0, 1]$.

Note that $[0, 1]$ isn't even always covered by the $I_k$. Consider the open intervals $\left(-1, \frac{1}{\sqrt{2}}\right), \left(\frac{1}{\sqrt{2}}, 2\right)$.

However, there are at most finitely many numbers that aren't covered by the union of the intervals. More precisely: Show that there are at most $n - 1$ numbers that aren't covered by the union of the $I_k$. Conclude that there are real numbers $x_1, \ldots, x_k$ so that each of the intervals $(x_i, x_{i + 1})$ is a subset of the union.