Measure zero and compact then content zero

Question

I'm trying to show that a compact set $E$ with measure zero has content zero. It seems simple because for every $\varepsilon$ I take a subcover of the rectangles, but the issue I'm facing here is that I can't take a subcover because the rectangles are closed. I tried using the interior but what guaranties that for every $\varepsilon > 0$ I'll have a set of rectangles such that $E \subset\bigcup_{k \in \Bbb{N}} R_k^\circ $ and $\sum_{k=1}^\infty \mu (R_k) < \varepsilon$.

Definition measure zero for every $\varepsilon >0$ we have a set of closed rectangles such that $E \subset\bigcup_{k \in \Bbb{N}} R_k$ and $\sum_{k=1}^\infty \mu (R_k) < \varepsilon$. Content zero is the same but with finitely many rectangles.

Note I don't know any measure theoretic concepts besides those.

Answer 1

Hint: a closed rectangle is contained in a slightly larger open rectangle.

For example, you can keep the same center and double the sidelengths. The new (open) rectangle has area $4$ times the area of the old (closed) rectangle. Choose a finite cover of those.