Lebesgue measure of compact and open sets Let $\lambda_n$ be the Lebesgue measure.
How to prove that 
1) For all compact sets $K \subset \mathbb{R^n}$ it's $\lambda_n(K)<\infty$
2) For all open sets $U \subset \mathbb{R^n}$ it's $\lambda_n(U)>0$
For 1) I thought about using that every compact set fits in a big enough cuboid with finite side length, but I don't see how it can be proved exactly.
2) Let $x \in U$. Then it exists a $\varepsilon>0$ such that $(x-\varepsilon,x+\varepsilon) \subset U$. 
So, $\lambda_n(x-\varepsilon,x+\varepsilon) \leq \lambda_n(U) \Rightarrow 0<2\varepsilon \leq \lambda_n(U) \Rightarrow \lambda_n(U)>0$
Can it be done like that or is there another way to show this?
 A: If $K$ is bounded there exists a number $M$ with $K \subset B(0,M)$. Since $B(0,M) \subset [-M,M]^n$ you get $$K \subset [-M,M]^n.$$
You don't need to know the Lebesgue measure of the cube. Rather, since the cube covers the bounded set $K$ the definition of the Lebesgue (outer) measure tells you that
$$\lambda_n(K) \le Vol_n([-M,M]^n) = (2M)^n.$$
Another way to prove the other result is by contradiction. Suppose that $U$ is an open set and that $\lambda_n(U) = 0$.  Since $U$ is open it contains a ball $B(x,r)$ so by monotonicity $\lambda_n(B(x,r)) = 0$ too.
If you don't know the Lebesgue measure of a ball this might not be enough to convince you of a contradiction, but using the translation invariance of the Lebesgue measure it implies $\lambda_n(B(0,r)) = 0$, and using the scaling property of the Lebesgue measure it implies in turn that $\lambda_n(B(0,2^k r)) = 2^k \lambda_n(B(0,r)) = 0$ for all $k$. These balls cover $\mathbb R^n$, so subadditivity gives you
$$\lambda_n(\mathbb R^n) \le \sum_k \lambda_n(B(0,2^k r)) = 0$$
which should be apparently false.
A: I try to answer to the first question by using the most general (read purely topological) definition of compactness: for the second question, it is my opinion, shared also by Yanko and Ricardo Freire, that you've already done a good job.
Definition Let $K$ be a subset of a topological space $(X,\mathscr{T})$: an open covering of $K$ is a family of open sets $\{C_i\}_{I\in I}$ such that
$$
K\subseteq\bigcup_{i\in I} C_i.
$$
An open covering $\{C_i\}_{I\in I}$ is said finite if it is made of a finite number of sets, i.e. $\sharp I\in \mathbb{N}$. 
$K$ is said to be compact if, from any of its open covering, it is possible to choose a finite subcovering.
Now, since $\mathbb{R}^n$ is covered by a numerable family of open cubes $\{Q_i\}_{I\in I}$ and since $K \Subset \mathbb{R}^n \iff$ $\{Q_i\}_{I\in I}$ covers also $K$, we can chose a finite subcovering $\{Q_i\}_{I\in I_0}\subset\{Q_i\}_{I\in I}$ such that
$$
K\subseteq \bigcup_{i\in I_0} Q_i \iff \lambda_n (K)\le \sum_{i\in I_0}\lambda_n (Q_i)<\infty
$$
