No, this isn't correct.
Chebyshev's inequality says that for any non-negative random variable $Z$ and any $\epsilon > 0$, we have $$P(Z \ge \epsilon) \le \frac{E[Z]}{\epsilon}.$$
You are applying this with $Z = |X_n - 1|$, so you get
$$P(|X_n - 1| \ge \epsilon) \le \frac{E[|X_n - 1|]}{\epsilon}.$$
Note the absolute value bars on the right side, which you apparently dropped. Now $E[|X_n - 1|] = 1/n$, not $-1/n$, and you get $P(|X_n - 1| \ge \epsilon) \le \frac{1}{n \epsilon}$. This is true but a bit silly, since you can get a better bound without Chebyshev: for $\epsilon \le 1$ you have $P(|X_n - 1| \ge \epsilon) = P(X_n = 0) = 1/n$, and for $\epsilon > 1$ you have $P(|X_n - 1| \ge \epsilon) = 0$. But this also doesn't help you apply Borel-Cantelli.
As an immediate sign that something was wrong, note that your argument "showed" that the probability of an event, which by definition is between 0 and 1, was less than or equal to a negative number. Uh oh.
In fact, you cannot prove from the given information that $X_n \to 1$ a.s., because that can be false. Suppose that the random variables $X_n$ were independent (the statement of the problem doesn't assume this, but also doesn't rule it out). Then you can use the second Borel-Cantelli lemma to show that $P(X_n = 0 \text{ i.o.}) = 1$ and also $P(X_n = 1 \text{ i.o.}) = 1$. Hence the sequence diverges almost surely.
It is true that $X_n \to 1$ in probability. You can use Chebyshev for this (if you use it correctly) but the "better bound" I mention above seems easier.
Best Answer
No, you cannot have such an inequality. Let $X_n$ take the values $0$ and $1$ with probabilities $\frac 1 {2^{n}}$ and $1-\frac 1 {2^{n}}$ respectively. Since $\sum P(X_n=0)<\infty$ Borel cantalli Lemma shows that $X_n \to 1$ almost surely. Now take $Y=\frac 1 4$. Then $P(Y\geq X_n) =\frac 1 {2^{n}}$ and $P(Y \geq \frac 1 2)=0$.