Suppose $S_1\subseteq S_2\subseteq \ldots \subset I=[0,1]$ is a such that each $I\setminus S_j$ contains a finite union of segments of total length $\varepsilon>0$. Then $\cup S_j\neq I$.
Here is a proof: By continuity of outer Lebesgue measure from below $S=\cup S_j$ has outer measure at most $1-\varepsilon$, and, in particular $S\neq I$.
I am interested in finding a proof that avoids measure theory and is as simple as possible.
Perhaps one can start by replacing $S_j$s by their closures $\Sigma_j$. Then one could apply the classification of closed sets in $I$ to say that each $I\setminus \Sigma_j$ is a countable union of intervals then do something with those intervals. Anything simpler?