I'm trying to verify below result, i.e.,
Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space and $X,Y:\Omega \to \mathbb R$ independent random variables. If $\mathbb E [ |X-Y|^2 ] < +\infty$ then both $X$ and $Y$ have finite second-moments.
Could you have a check on my below attempt?
Proof We define the map $f:\mathbb R \to \mathbb R \cup \{+\infty\}$ by $$ f(x) := \mathbb E [ |x-Y|^2 ]. $$
Then $f$ is Borel measurable. Because $X$ and $Y$ are independent, $$ \mathbb E[f(X)] = \mathbb E [ |X-Y|^2 ]. $$
If $\mathbb E [ |Y|^2] = +\infty$ then $f(x) = +\infty$ for all $x \in \mathbb R$ and thus $\mathbb E[f(X)] = +\infty$, which is a contradiction. This completes the proof.
It seems you are trying to do the following. Because $X$ and $Y$ are independent, $$ \mathbf{E}(f(X,Y) \mid X = x ) = \mathbf{E}(f(x,Y)), $$ therefore $$ \mathbf{E}(f(X,Y)) = \mathbf{E}(\mathbf{E}(f(X,Y) \mid X)) = \int dF_X(x) \mathbf{E}(f(x,Y)). $$ The exercise uses $f(u,v) = |u - v|^2,$ and the hypothesis implies at once that $\mathbf{E}(f(x,Y)) < \infty$ for almost every $x$ according to $F_X.$ One can conclude then that there exists $x$ such that $\mathbf{E}(f(x,Y)) < \infty.$ Thus, you should prove the following:
Claim: A necessary and sufficient condition for $Y$ to have second moment is that there exists $a \in \mathbf{R}$ such that $a - Y$ also have second moment. In other words, $\mathbf{E}(Y^2) < \infty$ if and only if there is some $a \in \mathbf{R}$ such that $\mathbf{E}(|a - Y|^2) < \infty.$
(My point being is that you are in the right track, except you haven't justified a few steps and your quantifiers are not quite right either.)