I've unaccepted kjetil's answer since, as was pointed out in a comment under Greenparker's answer, I believe they assume $X$ and $Y$ are independent.

The following answer should work when $X$ and $Y$ are dependent, by using whuber's suggestion:

\begin{align} \text{Var}(XY) &= E((XY)^2) - E(XY)^2 \\ &\le E(X^2Y^2) \\ &\le E(X^2)\sup(Y^2) \\ &= \text{Var}(X) + E(X)^2 \\ &< \infty \end{align}

where the final inequality follows from the mean and variance of $X$ being finite.

Note that the result also holds for any bounded $Y$ (since $\sup(Y^2)$ will be finite).

I've unaccepted kjetil's answer since, as was pointed out in the comments, it assumes $X$ and $Y$ are independent.

The following answer should work when $X$ and $Y$ are dependent, by using whuber's suggestion:

\begin{align} \text{Var}(XY) &= E((XY)^2) - E(XY)^2 \\ &\le E(X^2Y^2) \\ &\le E(X^2)\sup(Y^2) \\ &= E(X^2) \\ &= \text{Var}(X) + E(X)^2 \\ &< \infty \end{align}

Note that the result also holds for any bounded $Y$ (since $\sup(Y^2)$ will be finite).

I've unaccepted kjetil's answer since, as was pointed out in the comments under Greenparker's answer, I believe they assume $X$ and $Y$ are independent.

The following answer should work when $X$ and $Y$ are dependent, by using whuber's suggestion:

\begin{align} \text{Var}(XY) &= E((XY)^2) - E(XY)^2 \\ &\le E(X^2Y^2) \\ &\le E(X^2)\sup(Y^2) \\ &= \text{Var}(X) + E(X)^2 \\ &< \infty \end{align}

where the final inequality follows from the mean and variance of $X$ being finite.

I've unaccepted kjetil's answer since, as was pointed out in a comment under Greenparker's answer, I believe they assume $X$ and $Y$ are independent.

The following answer should work when $X$ and $Y$ are dependent, by using whuber's suggestion:

\begin{align} \text{Var}(XY) &= E((XY)^2) - E(XY)^2 \\ &\le E(X^2Y^2) \\ &\le E(X^2)\sup(Y^2) \\ &= \text{Var}(X) + E(X)^2 \\ &< \infty \end{align}

where the final inequality follows from the mean and variance of $X$ being finite.

Note that the result also holds for any bounded $Y$ (since $\sup(Y^2)$ will be finite).