Why are Haar measures finite on compact sets?
A Haar measure is finite on compact sets by definition: A Borel measure $\mu$ on a topological group $G$ is a (left) Haar measure if
$\mu$ is regular.
$\mu$ is left-invariant, i.e. $\mu(xU) = \mu(U)$ for any measurable $U \subset G$.
$\mu(U) > 0$ for all $U \subset G$ open.
$\mu(K) < \infty$ for all $K \subset H$ compact.
It is in fact possible to deduce a Haar measure is finite on compact sets from the rest of the axioms, if you assume there is some nonempty subset of finite measure.
Let $A\subseteq G$ such that $\mu(A)<\infty$. By regularity there is some open set $U\supseteq A$ such that $\mu(U)\leqslant\mu(A)+\varepsilon<\infty$ (for every $\varepsilon>0$, say $\varepsilon=1$). Since $U$ is open we also have $\mu(U)>0$. Now, pick some $u_0\in U$ and let $V:=u_0^{-1}U$; by left invariance $\mu(V)=\mu(U)$. We conclude that $V$ is open, $1\in V$ and $0<\mu(V)<\infty$.
The rest you've already showed: if $K$ is compact, then $\{kV\}_{k\in K}$ is an open cover of $K$, and hence there are $k_1,...,k_n$ such that $K\subseteq \bigcup_{i=1}^n k_iV$, and therefore $\mu(K)\leqslant n\cdot \mu(V)<\infty$.
It is not clear to me why some people put this in the definition of a Haar measure. Incidentally, it is also not necessary to assume a Haar measure is positive on open sets if you assume it is non-zero: if you had some open set $V\subseteq G$ such that $\mu(V)=0$, then clearly you'd have that $\mu(K)=0$ for every compact $K$, which means by regularity that $\mu(G)=0$.