I've seen this version of the Borel-Cantelli Lemma in Stein and Shakarchi:
Let $(E_k)_{k=1}^{\infty}$ be a countable family of measurable subsets of $\mathbb{R}^d$ satisfying $$\sum_{k=1}^{\infty}m(E_k) < \infty$$
Let $E$ be the set of $x\in \mathbb{R}^d$ such that $x\in E_k$ for infinitely many $k$.
Then $E$ is measurable and $m(E) = 0$.
Now in proving this it seems straightforward enough to show that the tail of the sequence goes to $0$, so for any $\varepsilon > 0$, choose $K\in\mathbb{N}$ such that $\sum_{k=K}^\infty m(A_k) < \varepsilon$.
Now by definition of $E$, we have $E\subset \bigcup_{k=K}^{\infty}E_k$ so by monotonicity and subadditivity of the exterior measure, $m_*(E) \leq \sum_{k=K}^\infty m(A_k) < \varepsilon$ which proves the result.
But did we need the measurability assumption? I did not use it anywhere in this argument, so is that a mistake on my part or would this hold for a collection of sets which may not all be measurable, but the series of their exterior measures still converges.
Yes, you can formulate this Borel-Cantelli lemma in terms of exterior measure (this would be then a slight generalization of the Borel-Cantelli lemma).
Let $(E_k)_{k=1}^{\infty}$ be a countable family of subsets (not necessarily Lebesgue measurable subsets) of $\mathbb{R}^d$ satisfying $$\sum_{k=1}^{\infty}m^*(E_k) < \infty$$
Let $E$ be the set of $x\in \mathbb{R}^d$ such that $x\in E_k$ for infinitely many $k$.
Then $E$ is Lebesgue measurable and $m(E) = 0$.
Proof: The proof goes as you suggested.
For any $\varepsilon > 0$, choose $K\in\mathbb{N}$ such that $\sum_{k=K}^\infty m^*(E_k) < \varepsilon$.
Now by definition of $E$, we have $E\subset \bigcup_{k=K}^{\infty}E_k$ so by monotonicity and subadditivity of the exterior measure, $$m^*(E) \leq m^* \left ( \bigcup_{k=K}^{\infty}E_k \right) \leq \sum_{k=K}^\infty m^*(E_k) < \varepsilon$$ which proves the result.