Proof verification: sequence a_n=(−1)^n does not converge

[CGandC]

Theorem: Show that the sequence $a_n = (-1)^n$ for all $n \in \mathbb{N},$ does not converge.
My Proof: Suppose that there exists a limit $L$ such that $a_n \rightarrow L$. Specifically, for $\epsilon = 1$ there exists $n_0$ s.t. for all $n > n_0$ then $|(-1)^n-L|<1$ , from this we can also infer that $|(-1)^{n+1}-L|<1$ for all $n > n_0$.
Let $n > n_0$ be arbitrary, then $2 = | (-1)^n - (-1)^{n+1} | \leq | (-1)^n - L | + | L - (-1)^{n+1} | < 1 + 1 = 2$, so since $n > n_0$ was arbitrary , hence for all $n>n_0$ we have 2<2 which is false, hence we get a contradiction.

My question: Regarding to when I wrote " for all $n>n_0$, we have 2<2 which is false, hence we get a contradiction. "
I'm trying to understand with respect to what there is a contradiction; we get that any statement of the form $\forall n>n_0. P(n)$ is false in my proof ( since we got $\forall n>n_0 . 2<2 )$ [ and $P(n)$ is a statement that depends on $n$ ], and since we already have in the proof the statement " for all $n > n_0$ then $|(-1)^n-L|<1$ " , then we get a contradiction. Is this correct?

I mean, in my proof I have $\forall n>n_0 . 2<2$ ( which is false ) and $\forall n>n_0.|(−1)^n+1−L|<1$ ( which is true by the assumption that there exists a limit ) , hence I get a contradiction and thus the assumption that the limit exists is false. So I wanted to know if my proof was correct.

 

[mfb]

The proof is fine.
If the series would converge then there needs to be an n0 such that 2<2 is true for all larger n. That's obviously wrong. 2<2 can't be true for any n.