$\newcommand{\E}{\operatorname{E}}\newcommand{\Var}{\operatorname{Var}}\newcommand{\Cov}{\operatorname{Cov}}$Let be $X$ and $Y$ two random variables, such that $\E[X^2]<\infty$ and $\E[Y^2]<\infty$, can we conclude that $\Var[aX+bY]$, with $a,b\in{\mathbb{R}}$ is finite?
My attemp: \begin{align} & \left|\Var[aX+bY]\right|=|a^2\Var[X]+b^2\Var[Y]+ab\Cov(X,Y)| \\[10pt] \leq {} & a^2\Var[X]+b^2\Var[Y]+|a||b| \left|\Cov(X,Y)\right| \\[10pt] \leq {} & a^2\Var[X]+b^2\Var[Y]+|a||b|\sqrt{(\Var[X])}\sqrt{(\Var[Y]})<+\infty \end{align}
So,in particular, we can conclude that $\E[(aX+bY)^2]<+\infty$
$\newcommand{\E}{\operatorname{E}}\newcommand{\Var}{\operatorname{Var}}\newcommand{\Cov}{\operatorname{Cov}}$You have $$\Var[X] = \E[X^2] - \E[X]^2 \le \E[X^2] < \infty$$ and $$\Var[a X + b Y] = a^2 \Var[X] + b^2 \Var[Y] + 2ab \Cov[X,Y]$$ and by the Cauchy-Schwarz inequality, you further have $$\Cov[X, Y]^2 \le \Var[X] \Var[Y]$$ and therefore finite variance for the linear combination.
You can also compute the expectation directly: $$\E[(aX+bY)^2] = \E[a^2 X^2 + b^2 Y^2 + 2ab \, X \, Y] \\= a^2 \E[X^2] + b^2 \E[Y^2] + 2ab \E[XY] \\\le a^2 \E[X^2] + b^2 \E[Y^2] + 2 \, |a| \, |b| \, |\E[XY]|$$ From the triangle inequality again you have $$|\E[XY]| \le |\Cov[X,Y]| + |\E[X]|\,|\E[Y]|$$ with the inequalities above and $\E[X]^2 \le \E[X^2]$ that becomes $$|\E[XY]| \le 2 \sqrt{\E[X^2]} \sqrt{\E[Y^2]}$$ and putting all together $$\E[(aX+bY)^2] \le a^2 \E[X^2] + b^2 \E[Y^2] + 2 \, |a| \, |b| \, \sqrt{\E[X^2]} \sqrt{\E[Y^2]} \\= \Big(|a| \sqrt{\E[X^2]} + |b| \sqrt{\E[Y^2]}\Big)^2$$ which is finite.