Show that $\operatorname{Var}{X} < \infty \iff \mathbb E[X^2] < \infty$
I am attempting to use the above without the fact that $\mathbb E[(X-\mathbb E[X])^2]=\mathbb E[X^2]-\mathbb E[X]^2$ if $\mathbb E[X] < \infty$, since we have not been given that $\mathbb E[X] < \infty$. (although do $\operatorname{Var} X < \infty$ as well as $\mathbb E[X^2]< \infty$ respectively imply that $\mathbb E[X] < \infty$ ?)
"$\Rightarrow$"
$\operatorname{Var} X= \mathbb E[(X-\mathbb E[X])^2]=\int (X-\mathbb E[X])^2dP=\int(X^2-2X\mathbb E[X]+\mathbb E[X]^2)dP$
Since $\operatorname{Var} X < \infty$, it is well-defined as an integral and therefore $\int(X^2-2X\mathbb E[X]+\mathbb E[X]^2)dP<\infty$ implies $\int X^2dP<\infty$ so $\mathbb E[X^2]<\infty$
"$\Leftarrow$" If $\mathbb E[X^2] < \infty$ implies that $\mathbb E[X] < \infty$ then it's easy game. Since $\operatorname{Var} X = \mathbb E[X^2]- \mathbb E[X]^2$ which is $< \infty$
Any pointers, improvements, pointing out of mistakes would be greatly appreciated.
To define variance as $E(X-EX)^{2}$ (which is the definition you have adapted) $EX$ must exist either as real number or $\pm \infty$. If $EX= \pm\infty$ then $(X-EX)^{2}=\infty$ almost surely so variance is $\infty$. Hence $var(X) <\infty$ implies $EX$ also exists as a finite number. This is the reason nobody defines variance of a random variable that does not even have finite mean. Once you know that $X$ has finite mean you can write $var(X)=EX^{2}-(EX)^{2}$ from which you can easily see that $EX^{2} <\infty$ if the variance is finite.