I came up with the following proof about what's stated in the title. I know it is wrong but can't find the error.
Let $X$ be a random variable with finite mean ($EX<\infty$). The function $f(t)=-t^2$ is concave over $\mathbb{R}$, so applying Jensen inequality we get $Ef(X)\leq f(EX)$, that is, $E(-X^2)\leq f(EX)$, so $-X^2$ has finite mean and that implies $E(|-X^2|)=E(X^2)<\infty$. Therefore $Var(X)=EX^2-(EX)^2$ is finite.
I know it must be wrong because Pareto distribution is a counterexample.
Any help is highly appreciated.
You've shown that the expectation of $-X^2$ is bounded above, but you haven't shown it to be bounded below. You'd need to show this to guarantee that it's finite.
Note that it's more easily bounded above by noting that $-X^2\le0\implies \mathbb{E}(-X^2)\le0$.