Let $A \subset \mathbb{R}^{d}$ ($d$ here is a natural number). The outer measure $\lambda_{*}(A)$ of $A$ is defined as follows:
If $\sigma\left(\left(Q_{n}\right)_{n \in \mathbb{N}}\right)=\infty$ for all coverings $\left(Q_{n}\right)_{n \in \mathbb{N}} \in \mathcal{C}(A)$, $\lambda_{*}(A):=\infty$.
Otherwise, $$ \lambda_{*}(A):=\inf \left\{\sigma\left(\left(Q_{n}\right)_{n \in \mathbb{N}}\right):\left(Q_{n}\right)_{n \in \mathbb{N}} \in \mathcal{C}(A), \sigma\left(\left(Q_{n}\right)_{n \in \mathbb{N}}\right)<\infty\right\} $$ I want to prove that $\lambda_{*}\left(\mathbb{R}^{d}\right)=\infty$ and $\lambda_{*}(\emptyset)=0$.
I tried to use the fact that $\mathbb{R}^{d}=\bigcup_{n=1}^{\infty}[-n, n]^{d}$, so it follows that every subset of $\mathbb{R}^{d}$ can be covered by a countable union of closed cubes:
$\lambda_{*}\left(\bigcup_{n=1}^{\infty}[-n, n]^{d}\right)$ $:=\sum_{n=1}^{\infty} 2 n^{d}:=\lim _{N \rightarrow \infty} \sum_{n=1}^{N} 2 n^{d}=\infty$ (since $d \geqslant 1$)
But I am not sure.
For the empty set $\forall A \subseteq \mathbb{R}^{d}$ $\phi \in A$, I thought o taking the following coverins $\bigcup_{n=1}^{\infty}\left[\frac{\varepsilon}{2^{n}}, \frac{\varepsilon}{2^{n}}\right]^{d}$, so that inf of the outer measure will be $0$ as $n$ tends to $\infty$.
First we can note that the outer measure doesn't change if we allow rectangles of any kind, even ones that aren't cubes and aren't closed since given a cover of $A$ by rectangles, we can create a cover of $A$ by closed cubes with arbitrarily small extra area.
To prove the results you want, first use Heine Borel to prove that the outer measure of any cube is it's volume, i.e. the product of it's side lengths (Heine Borel is unavoidable here, the proposition fails if we use $\mathbb{Q}^d$ instead of $\mathbb{R}^d$). This will take some labor.
Now by monotonicity of the outer measure, for any $n \in \mathbb{N}$ we have $(2n)^d = \lambda^*([-n, n]^d) \leq \lambda^*(\mathbb{R}^d)$. Letting $n \to \infty$ gives $\infty \leq \lambda^*(\mathbb{R}^d)$, so $\lambda^*(\mathbb{R}^d) = \infty$. $\emptyset$ is the rectangle $(0, 0)^d$, so $\lambda^*(\emptyset) = (0 - 0)^d = 0$. Your covering idea seems right, but the execution is not there. Let $\varepsilon > 0$ be arbitrary. We have $(-\varepsilon, \varepsilon)^d \supset \emptyset$. Thus by definition of the outer measure, $\lambda^*(\emptyset) \leq (2\varepsilon)^d$. Letting $\varepsilon \to 0$ gives $\lambda^*(\emptyset) \leq 0$. Thus $\lambda^*(\emptyset) = 0$.