Showing two measures are equal up to scalar multiplication.

It helps to know that finite Borel measures on a metric space $X$ are regular. In particular, if $E \subset X$ is a Borel set, $\mu$ is a finite Borel measure, and $\epsilon > 0$, then there exists an open set $G \supset E$ with $\mu(G) < \mu(E) + \epsilon$.

Assume that the stated conditions on $\mu$ and $\nu$ hold (and in particular $\nu(B(x,r)) > 0$ for all $x \in X$ and $r > 0$).

Let $E \subset X$ be a Borel set. Fix a constant $t > 1$.

For every $x \in E$ there exists $r_x > 0$ with the property that $0 < r < r_x$ implies $$\frac ct < \frac{\mu(B(x,r))}{\nu(B(x,r))} < ct,$$ which can be expressed as $$0 < r < r_x \implies \frac ct \nu(B(x,r)) \le \mu(B(x,r)) \le ct \nu(B(x,r)).$$

Let $\epsilon > 0$ and let $G \supset E$ be an open set with $\nu(G) < \nu(E) + \epsilon$. Define a Vitali cover ${\cal V}$ of $E$ by $$ {\cal V} = \{B(x,r) : x \in E,\ 0 < r < r_x,\ B(x,r) \subset G\}.$$ According to Vitali's covering theorem there exists a countable disjointed subfamily ${\cal V}'$ of $\cal V$ satisfying $$\mu(E) \le \mu \left( E \setminus \bigcup_{B \in {\cal V}'} B \right) + \mu \left(\bigcup_{B \in {\cal V}'} B \right) \le \sum_{B \in {\cal V}'} \mu(B) \le ct\sum_{B \in {\cal V}'} \nu(B)$$ Since the family ${\cal V}'$ is pairwise disjoint and each $B \in {\cal V}'$ is a subset of $G$ you have in addition that $$\sum_{B \in {\cal V}'} \nu(B) \le \nu(G) <\nu(E) + \epsilon.$$ Thus $\mu(E) < ct(\nu(E) + \epsilon)$ and by letting $\epsilon \to 0^+$ you get $$\mu(E) \le ct \nu(E).$$

Now let $\epsilon > 0$ and let $H \supset E$ be an open set with $\mu(H) < \mu(E) + \epsilon$. Repeat the steps above to find that $$\frac ct \nu(E) \le \mu(E)$$ and conclude $$\frac ct \nu (E) \le \mu(E) \le ct \nu(E).$$ Finally let $t \to 1^+$.