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.
Suppose that $X_n \to X$ almost surely. Thus if $\Omega_0 = \{X_n \to X\}$ then $\mathbb P(\Omega_0) = 1$. Then for any fixed $c \in \mathbb R$
$$ \Omega_0 \cap \{X < c\} = \Omega_0 \cap\{X_n < c\ \text{infinitely often}\}$$
Then apply Borel-Cantelli to justify that $\{X_n < c\ \text{infinitely often}\}$ happens with probability either 0 or 1, thus
$$\mathbb P(X < c) = \mathbb P(\Omega_0 \cap \{X < c\}) = \mathbb P(\Omega _0\cap \{X_n < c\ \text{i.o.}\}) = \mathbb P(\text{$X_n < c$ i.o.})$$ and will equal 0 or 1. Use this to deduce that $X$ is almost surely constant.
Best Answer
$X=\lim \{X_{k+1},X_{k+2},\cdots\}$ so $X$ is independent of $X_1,X_2,...,X_k$. This is true for each $k$. Hence $X$ is independent of the entire sequence $\{X_n\}$. But, being the limit of this sequence, it is measurable w.r.t. $\sigma (X_1,X_2,...)$. Hence $X$ is independent of itself. This implies that $X$ is a constant.
[Let $F(X)=P\{X\leq x\}$. Since $\{X\leq x\}=\{X\leq x\}\cap \{X\leq x\}$ we get $F(X)=F(x)F(x)$. Hence $F(x)=0$ or $1$ for each $x$. If $c =\sup \{x:F(x)=0\}$ then $F(x)=0$ for $x <c$ and $1$ for $x \geq c$. This implies that $P\{X\neq c\}=0$ so $X=c$ almost surely].