Defining Measure 0 Sets with Open or Closed Rectangles

I have been reading through Calculus on Manifolds by Michael Spivak, and I am not understanding what he states after defining measure 0 sets. On page 50, he states

"A subset $$A$$ of $$\mathbb{R}^n$$ has (n-dimensional) measure 0 if for every $$\varepsilon > 0$$ there is a cover $$\{U_1, U_2, U_3, \dotsc\}$$ of $$A$$ by closed rectangles such that $$\sum_{i = 1}^\infty v(U_i) < \varepsilon$$."

And follows this by stating

"The reader may verify that open rectangles may be used instead of closed rectangles in the definition of measure 0."

I have been trying to prove the equivalence between the definition of measure zero with open rectangles and closed rectangles. I have been able to prove that if we can do this with open rectangles, then we can do it with closed rectangles. However, I have not been able to prove the other direction. I have written the question as the following statement:

Suppose that for $$A \subset \mathbb{R}^n$$ and $$\varepsilon > 0$$, there is a cover $$\{F_1, F_2, F_3, \dotsc\}$$ of $$A$$ by closed rectangles such that $$\sum_{i = 1}^\infty v(F_i) < \varepsilon$$. Then, for any $$\varepsilon' > 0$$, there is a cover $$\{U_1, U_2, U_3, \dotsc\}$$ of $$A$$ by open rectangles such that $$\sum_{i = 1}^\infty v(U_i) < \varepsilon'$$.

Spivak Page 50

I included the Lebesgue measure tag, since I understand Spivak's definition of volume of a rectangle to be connected, but if this is incorrect, feel free to remove that tag.

Suppose that for $$A\subset\mathbb{R}^n$$ and $$\varepsilon>0$$, there is a cover $$\{F_1, F_2, F_3, \dots\}$$ of $$A$$ by closed rectangles such that $$\sum_{i = 1}^\infty v(F_i) < \varepsilon$$. Take $$\varepsilon' > 0$$, and consider a cover $$\{F_1, F_2, F_3, \dots\}$$ of $$A$$ by closed rectangles such that $$\sum_{i = 1}^\infty v(F_i) < \frac{\varepsilon'}2$$. Replace each $$F_i$$ by a larger open rectangle $$U_i$$ such that $$v(U_i). Then $$\{U_1,U_2,U_3,\dots\}$$ is a cover of $$A$$, and $$\sum_{i=1}^\infty v(F_i) <\frac{\varepsilon'}2+\frac{\varepsilon'}2=\varepsilon'$$.
If $$F_k=[a_{k1},b_{k1}]\times [a_{k2},b_{k2}] \times ... \times [a_{kn},b_{kn}]$$ take $$U_k=(a_{k1}-\frac {\epsilon} {2^{k}},b_{k1}+\frac {\epsilon} {2^{k}})\times ... \times (a_{kn}-\frac {\epsilon} {2^{k}},b_{kn}+\frac {\epsilon} {2^{k}})$$.