Prove both the exterior measure of both a point and $\emptyset$ is zero. This is an attempt of that proof.
Please also check this -- my definition of exterior measure -- which I believe is: the exterior measure of a set $E$, denoted $m_*\left(E\right)$, is defined as
$$m_*(E) = \inf\sum_{j=1}^{+\infty} \left|Q_j\right|$$
where the infimum is taken over all coverings of $E$ by almost disjoint cubes, described by $E \subset \bigcup_{j=1}^{+\infty} Q_j$.
proof
Let $x \in \mathbb{R}^d$ be a point. Define $Q_1 \subset \mathbb{R}^d$ to be be the cube centered around (centered on side lengths) the point $x$ with side length equal to one. Here $E = \{x\} \subset \bigcup_{j=1}^{+\infty} Q_j = Q$, where $Q_2, Q_3, \dots = \emptyset$. So the volume of $Q$ which equals $\sum_{j=1}^{+\infty} \left|Q_j\right|$, is clearly $1$. But we can construct an infinite sequence of such coverings $Q$, the next one with $Q_1$ having side length $1/2$, then the next one $1/4$, and so on. Clearly, the infimum of such a sequence is zero, and clearly $\{x\}$ is in each $Q$. Therefore
$$m_*\left(x \in \mathbb{R}^d\right) = 0$$
Now, to prove the empty set has the same exterior measure, it is enough to see that $\emptyset \subset \{x\}$ is covered by the same $Q$'s, and therefore has exterior measure of zero.
Thanks for the feedback.
The proof looks correct, although a little overcomplicated. Note that the exterior measure is defined by the infimum over all covers of rectangles. So to prove that $m_*(E) = 0$, you have to prove that for any $\epsilon > 0$ you can find an open cover of rectangles, $\{Q_k\}$, such that $\sum\limits_{k=1}^\infty |Q_k| < \epsilon$. If the point is in $\mathbb{R}^d$, you can just construct a cube around the point $p$ with side length $\left( \frac{\epsilon}{2}\right)^\frac{1}{d}$ and note that this cube has volume $\frac{\epsilon}{2}$.