If two Borel measures coincide on all open sets, are they equal?

Your guess is correct : the answer is "No" in general;

For example, let $\mu_1$ be the counting measure on $\mathbb R$, and let $\mu_2$ be the measure defined by $\mu_2(\emptyset)=0$ and $\mu_2(A)=\infty$ if $A\neq\emptyset$.

On the other hand, if the space $X$ is the union of an increasing sequence of open sets on which the two measures are finite, the the answer is "Yes". This follows from the monotone class theorem.

Here is an alternative, perhaps slightly easier proof using Dynkin's π−λπ−λ Theorem: https://math.stackexchange.com/a/813414/283164 Although it is sketched for $\mathbb R$, it works more generally.

BTW, Lemma 7.1.2. (p. 68) of Measure Theory, volume 1, Vladimir I. Bogachev:
If two finite signed Borel measures on any topological space coincide on all open sets, they coincide on all Borel sets.

Its simple proof uses:
Lemma 1.9.4. If two probability measures on a measurable space $(X,A)$ coincide on some class $E\subset A$ that is closed with respect to finite intersections, then they coincide on the $\sigma$-algebra generated by $E$.

Link to Lemma 7.1.2

Hint: what's the collection of all set where the 2 measures agree? Does that collection form a monotone class? Does it contains all open set?