$\sum_{k=1}^{\infty} X_k < \infty \to \sum_{k=1}^{\infty} VarX_k < \infty$

Let $(X_k)$ be independent random variables, with $P(X_k \le M)=1$, i.e bounded by a constant M. Proe that if $\sum_{k=1}^{\infty} X_k < \infty$ and for $k \in \mathbb{N}$$EX_k = 0$ then$\sum_{k=1}^{\infty} VarX_k < \infty$.

My attempt:

$\sum_{k=1}^{\infty} VarX_k = \sum_{k=1}^{\infty} (EX_k^2 - (EX_k)^2) = \sum_{k=1}^{\infty} (EX_k^2)$.Now for sufficiently large $k$ say $k \ge N$ there exists $\delta$, $\epsilon > 0$ s.t. $P(|X_k^2| \ge \epsilon) < \delta$. We can now bound the expectation. We define $A = \{|X_k^2| \ge \epsilon\}$. $EX_k^2 = \int_{A} X_k^2 \, dP + \int_{\Omega \setminus A} X_k^2 \, dP \le \epsilon(1-\delta) + M^2\delta .$ As $\epsilon, \delta$ approach $0$ we get the right side to approach $0$(In fact we do not because as $\epsilon \to 0$ we get right side to approach $M^2$). So we get that for large $n,m$$\sum_{k=n}^{m} X_k$ to be small, hence proof is completed via Cauchy criterion for series is a convergence.

Edit:Ok now I see that a proof is flawed because as $\epsilon \to 0$ we have $\delta \to 1$ so the quantity $M^2\delta$ doesn't become small. Is there a way to save this proof?

Edit 2: Maybe the idea is correct? As $\sum_{k=1}^{\infty} X_k$ converges we have $X_k \to 0$ and as $X_k$'s are bounded by $M$ then for arbitrarily small $\epsilon , \delta > 0 $ we can find $N$ s.t. for all $k \ge N$$P(X_k^2 \ge \epsilon) < \delta$. So we can manipulate both $\epsilon$ and $\delta$ and so Cauchy criterion for series convergence holds.