The Theorem is as follows: If $\delta > 0$ and $E \subseteq \mathbb{R}^d$ is measurable then $\delta E$ is measurable and $m(\delta E) = \delta^dm(E)$.
My proof is as follows:
Since $E$ is measurable, for all $\epsilon > 0$ there is an open set $O$ where $E \subseteq O$ such that $m_*(O - E) \leq \epsilon$. For $\delta E$ we can choose $\delta O$. We now have $\delta E \subseteq \delta O$ and $m_*(\delta O - \delta E) \leq \epsilon$ so $\delta E$ is measurable (I think I may have skipped over some reasoning here.)
Next to show $m(\delta E) = \delta^dm(E)$ we can write $E$ as a countable union of cubes $\bigcup_{j=1}^\infty Q_j$ where each $Q_j \subseteq \mathbb{R}^d$ is a cube. So this means we can form a covering of $\delta E$ by elongating the $d$ sides of each $Q_j$ by a factor of $\delta$. This would mean $$\delta E \subseteq \bigcup_{j = 1}^\infty \delta^dQ_j$$ Since $E$ is measurable we can now perform the following: \begin{align*} m(\delta E) &= \inf \sum_{j = 1}^\infty |\delta Q_j|\\ &= \delta^d\inf \sum_{j = 1}^\infty |Q_j|\\ &= \delta^d m(E) \end{align*} Could I please check if my proof is correct and that I haven't missed any steps? I'm still new to measure theory so I appreciate your help!
Your proof is not completely valid, but the ideas are good.
You should probably first prove that $m_*(\delta G) = \delta^dm_*(G)$ for any set $G$ so that you can then say that $m_*(\delta O - \delta E) \leq \delta^m\epsilon$ (in your proof of measurability of $\delta E$).
Your proof of the equality $m(\delta E) = \delta^dm(E)$ is not completely clear to me either.
First you decompose $E$ into cubes, which is not the way I would go. As I said we need to prove $m_*(\delta G) = \delta^dm_*(G)$ for any set $G$. So I would rather say that $m_*(G) = \inf \sum_{j = 1}^\infty |Q_j|$ where the infimum is taken over all countable coverings of $G$ with cubes.
Now you say that $m_*(\delta E) = \inf \sum_{j = 1}^\infty |\delta Q_j|$, but it is not clear to me. Think about it - what is the infimum even taken over assuming the cubes are a decomposition of $E$ (like in your original solution)? Now assuming that we are considering all coverings of $E$ instead of decompositions it makes more sense and is indeed true (but we have to prove it).
So now we will prove $m_*(\delta G) = \inf \sum_{j = 1}^\infty |\delta Q_j|$. The "elongated" covering is derived from some covering of $G$; a priori we don't know if every covering of $\delta G$ is derived in this way. So all we can really say is that $$m_*(\delta G) = \inf \sum_{j = 1}^\infty |P_j| \leq \inf \sum_{j = 1}^\infty |\delta Q_j| = \delta^d\inf \sum_{j = 1}^\infty |Q_j| = \delta^d m_*(G)$$ where the infimum with $P_j$ is taken over all countable coverings of $\delta G$ with cubes, and the infima with $Q_j$ are taken over all countable coverings of $G$ with cubes (notice the difference!). The inequality holds because an infimum over a bigger family of coverings is not bigger than an infimum over a smaller family of coverings.
What I mean is: $\delta Q_i$ is a covering of $\delta G$, but a priori we don't know if every covering $P_i$ of $\delta G$ is necessarly of the form $\delta Q_i$ for some covering $Q_i$ of $G$. Thus all coverings of $\delta G$ form a (not necessarly strictly) bigger family (in the sense of inclusion) than the "elongated" coverings derived from coverings of $G$.
Thus $m_*(\delta G) \leq \delta ^d m_*(G)$. Now do the same for $\delta^{-1}$ in place of $\delta$ and $\delta G$ in place of $G$ to get the $\geq$ inequality.