Show that if X and Y are i.i.d random variables such that $\mathbb{E}(X-Y)^{2}<\infty$, then $\mathbb{E}X^{2}<\infty$

According to Feller, An Introduction to Probability Theory, vol 2, second edition, Lemma 3 on page 151, if $X$ and $Y$ are independent random variables, $E|X+Y|^\alpha$ exists if and only if both $E|X|^\alpha$ and $E|Y|^\alpha$ exist.

This, with the notational change that Feller's $Y$ is the OP's $-Y$, answers the original question without any extra condition.

Here is the argument, cut down to the problem at hand. Since for any fixed $a$ the moment $E|X-a|^2$ exists if and only if $E|X|^2$ exists, we may assume, without loss of generality, that $P(X\ge0)\ge 1/2$ and $P(X\le0)\ge1/2.$ (So that $0$ is a "median" for $X$.) If either of $X\le0, Y>t$ or $X\ge0, Y<-t)$ occurs then $|X-Y|>t$, so $$P(|X-Y|>t)\ge P(X\le0)P(Y>t) + P(X\ge0)P(Y<-t)\ge\frac 1 2 P(|Y|>t).$$ Now use the formula $E|T|^2=\int_0^\infty 2t P(|T|>t) dt$ on the extreme sides of the displayed inequality.

Here is a small twist on Kimchi lover's answer, which is slightly more direct. Using the same reduction, one may assume without loss of generality that $0$ is a median of $X$ (and thus of $Y$). Then:

$$\mathbb{E} ((X-Y)^2) \geq \mathbb{E} (X^2 \mathbf{1_{X \geq 0}} \mathbf{1_{Y \leq 0}}) + \mathbb{E} (X^2 \mathbf{1_{X \leq 0}} \mathbf{1_{Y \geq 0}}).$$

By independence,

$$\mathbb{E} ((X-Y)^2) \geq \mathbb{E} (X^2 \mathbf{1_{X \geq 0}}) \mathbb{P} (Y \leq 0) + \mathbb{E} (X^2 \mathbf{1_{X \leq 0}}) \mathbb{P} (Y \geq 0) \geq \frac{\mathbb{E} (X^2 \mathbf{1_{X \geq 0}})+ \mathbb{E} (X^2 \mathbf{1_{X \leq 0}})}{2} = \frac{\mathbb{E} (X^2)}{2}.$$

This avoids the formula relating the moments and the tails of $X$.