Independent random variables, equivalent in distribution, almost sure convergence

I am struggling with forward direction of this proof:

Suppose $\{X_n, n\geq 1 \}$ are independent random variables such that $X_n \overset{d}{=} Y_n$ for all $n \geq 1$. Suppose further that, for all $n \geq 1$, there is a constant $K$ such that $$|X_n| \vee |Y_n| \leq K.$$

Then $\sum_n (X_n - Y_n)$ converges almost surely if and only if $\sum_n \mathrm{Var}(X_n) < \infty$.

I was able to use the Kolmogorov Convergence Criterion to show that $\sum_n (X_n - Y_n)$ converges almost surely if $\sum_n \mathrm{Var}(X_n) < \infty$. For the forward direction, I was thinking about using the Kolmogorov Three Series Theorem,

Let $\{X_n, n \geq 1\}$ be an independent sequence of random variables. Then $\sum_n X_n$ converges almost surely if and only if there exists $c > 0$ such that

$(i) \sum_n P[|X_n| > c] < \infty$

$(ii) \sum_n Var(X_n 1_{[|X_n| \leq c]}) < \infty$

$(iii) \sum_n E(X_n 1_{[|X_n| \leq c]}) < \infty$

However, I was not sure how to make use of the assumption that $|X_n| \vee |Y_n| \leq K$ in the context of the $c$ mentioned in the three series theorem. If we choose $c = K$, then (ii) becomes

$$\sum_n Var(X_n) < \infty$$

but this is what we want to show, so I don't think this is the right approach.