I have a problem with the next proof. Sorry for my bad english, but this is not my language.
Let $\mathcal{P}(\mathbb{R}^n)$, the power set of $\mathbb{R}^n$, and $\mathfrak{C}\subset\mathcal{P}(\mathbb{R}^n)$, where $\mathfrak{C}$ is the set of compact sets of $\mathbb{R}^n$, we define the next operations with $\lambda\in\mathbb{R}$ and $A,B\in\mathfrak{C}$:
$$A+B=A\cup B \ \ \text{and} \ \ \lambda A=\{\lambda x:x\in A\}$$
I need proof the properties of vector space with this operations. In the sum, four properties are true (the inverses doesn't exist). But, the problem is with the scalar closure. My idea is apply the definition with open covers, but, really, I don't know how.
Anyway, I thought that is not true, but, I can't find the counterexample.
If I understand the question correctly you are asked to check all defining properties in the definition of a vector spaces and see which one are false and which one are true.
And I guess your problem is to verify the following:
$\lambda \in \mathbb{R}$ and $A \in \mathfrak{C}$ $\quad \Rightarrow \quad \lambda \cdot A \in \mathfrak{C}$ ?
If so, then the answer was already in the comment by @DonAntonio:
(I assume you work with the euclidean topology) If you write $\lambda \colon \mathbb{R}^n \to \mathbb{R}^n$, where $x \mapsto \lambda x$, then your scalar multiplication on $\mathfrak{C}$, $\lambda \cdot A$, is just the image of $A$ under the map $\lambda$, i.e $\lambda \cdot A = \lambda(A)$. But it is easy to show (first year proof I guess in an undergraduate analysis course) that the map $\lambda$ is countinuous (even a homeomorphism if $\lambda \neq 0$). And again it is a short proof to show that continuous maps map compact sets to compact sets. So since $A$ is compact so is $\lambda(A)=\lambda \cdot A$.