Contradiction proof that a sequence cannot converge in probability

Let $Y$ be distributed as a $N(0,1)$ random variable. Let $(Y_{n})_{n\in\mathbb{N}}$ be the sequence of random variables such that $Y_n=(-1)^nY.$ Prove that $Y_n$ does not converge in probability to any random variable.

Attempt: Assume $Y_n\rightarrow^pW$ for some random variable $W$. Then we have that $\lim_{n\rightarrow\infty}P(|Y_n-W|\le\varepsilon)=1$. However, I don't know how to apply this here. Perhaps proof by contradiction?--

Note that $|Y|=|Y_n|$ for all $n$.

If $|Y(\omega)| = |Y_n(\omega)| > 2\epsilon$ and $|Y_n(\omega) - W(\omega)| \le \epsilon$, then $|Y_{n+1}(\omega) - W(\omega)| = |-Y_n(\omega) - W(\omega)| > \epsilon$. Thus, $$P(|Y| > 2\epsilon, |Y_n - W| \le \epsilon) \le P(|Y_{n+1} - W| > \epsilon).$$

If $Y_n \overset{p}{\to} W$, then the right-hand side tends to zero as $n \to \infty$, and therefore so does the left-hand side.

Then, $$P(|Y| \le 2\epsilon, |Y_n - W| \le \epsilon) = P(|Y_n-W| \le \epsilon) - P(|Y| > 2\epsilon, |Y_n-W| \le \epsilon) \overset{n \to \infty}{\longrightarrow} 1-0 = 1.$$

This implies $P(|Y| \le 2\epsilon) = 1$, which is false for $Y \sim N(0,1)$. (More generally, the only way $Y_n = (-1)^n Y$ converges in probability is if $Y=0$ almost surely.)