My proof of divergence of $(-1)^n$ Should I shorten my proof?
  (Also, should I try to prove without contradiction?)
We consider the sequence
  $(x_n)_{n \in \mathbb{N}}$,
  where
  $x_n = (-1)^n$.
$\textbf{Lemma.}$
  For every element $x_n$ of the sequence $(x_n)$,
  we have $|x_n| = 1$.
  (We could prove this by induction on $n$.)
$\textbf{Theorem.}$
  $(x_n)$ diverges.
$\textit{Proof.}$
  We prove the theorem by contradiction.
  To that end,
  we assume that $(x_n)$ is not divergent, i.e. we assume that it is convergent.
  With that said, we are done as soon as a contradiction is deduced.
  By assumption, there is an $x \in \mathbb{R}$ such that
  \begin{equation*}
    \forall \varepsilon \in \mathbb{R}, \varepsilon > 0 :
    \exists N           \in \mathbb{N}                  :
    \forall n           \in \mathbb{N}, n > N           :
    |x_n - x| < \varepsilon .
  \end{equation*}
  We choose $\varepsilon = 1$.
  By assumption, there is an $N \in \mathbb{N}$ such that
  \begin{equation*}
    \forall n \in \mathbb{N}, n > N :
    |x_n - x| < 1 .
  \end{equation*}
  We choose $n = N + 1$.
  Hence, both $|x_n - x| < 1$ and $|x_{n + 1} - x| < 1$.
  Thus,
  \begin{equation*}
    |x_{n + 1} - x| + |x_n - x| < 2 .
  \end{equation*}
  Moreover,
  \begin{equation*}
    \begin{split}
      2 & =   |2| \\
        & =   |2| \cdot  1  \\
        & =   |2| \cdot |x_{n + 1}| && | \text{ by Lemma} \\
        & =   |2 x_{n + 1}| && | \text{ by multiplicativeness of abs. val.} \\
        & =   |x_{n + 1} + x_{n + 1}| \\
        & =   |x_{n + 1} + (-1)x_{n}| \\
        & =   |x_{n + 1} - x_{n}| \\
        & =   |x_{n + 1} + 0 - x_{n}| \\
        & =   |x_{n + 1} + (-x + x) - x_{n}| \\
        & =   |(x_{n + 1} - x) + (x - x_{n})| \\
        & \le |x_{n + 1} - x| + |x - x_{n}| && | \text{ by subadditivity of abs. val.} \\
        & =   |x_{n + 1} - x| + |x_{n} - x| \qquad && | \text{ by evenness of abs. val.} \\
    \end{split}
  \end{equation*}
  Hence, by transitivity, we have $2 < 2$.
  Obviously, we deduced a contradiction. QED
 A: You can use the following theorem: "If a sequence converges, then every subsequence converges to the same limit".
A: I would shorten the proof.
Pick any $L \in \mathbb{R}$.
If $L \ge 0$, then we have $|x_n-L| \ge 1$ for all odd $n$. Hence $x_n$ does not converge to $L$.
If $L < 0$, then we have $|x_n-L| \ge 1$ for all even $n$. Hence $x_n$ does not converge to $L$.
A: The shortest proof of this would be to show that the sequence is not Cauchy, therefore cannot converge.   Let $\epsilon =1$.  Assume for contradiction sake that the sequence is cauchy, therefore there exists an $N$ such that for all $m\ge N,n\ge N,|a_n-a_m|<1$.  Thus, taking $n=N,m=N+1$, we have $|(-1)^N-(-1)^{N+1}|<1$,  but this is false, since this number is actually 2 (Since $N$ and $N+1$ have opposite parity).  Therefore it's not cauchy, hence it does not converge
A: I do not know how much you know about sequences but the shortest proof of the divergence of this sequence is that the two subsequences $x_{2n}$ and $x_{2n+1}$ that are constant therefore convergent do not converge to the same limit
A: I think there are better ways to prove it.

Proposition A: If $x_n$ converges, then the sequence of differences $x_{n+1}{-}x_n$ converges to $0$.
Proposition B: If $x_n$ has subsequences $a_n$, $b_n$ which converge to different limits, then $x_n$ doesn't converge.

Either of these two immediately proves it and their proofs are no harder than what you are trying to do.
Take e.g. the first: For all $\varepsilon>0$ there's $N$ such that for all $n\ge N$, $|x_n-L|<\varepsilon/2$, so for all $n\ge N$, $$|x_{n+1}-x_n|=|(x_{n+1}-L)-(x_n-L)|\le |x_{n+1}-L|+|x_n-L|\le\varepsilon/2+\varepsilon/2=\varepsilon.$$
Done!
A: To give yet another answer - for any $n$, we have $\sup_{k\geqslant n} x_n=1$, so that $\limsup_{n\to\infty}x_n = 1$, while $\inf_{k\geqslant n} x_n=-1$, so that $\liminf_{n\to\infty}x_n=-1$. Since $\limsup_{n\to\infty}x_n\neq\liminf_{n\to\infty} x_n$, the sequence does not converge.
A: We shall assume that the sequence $\{x_n\}$ converges and reach a contradiction. Let $L$ denote the number the sequence converges to, and set $\hat{\varepsilon}=\frac{1}{2}$. By definition there is $\hat{n} \in \mathbb{N}$ so that $|x_n-L|<\hat{\varepsilon}$ whenever $n \geq \hat{n}$. Hence $|x_{\hat{n}}-x_{\hat{n}+1}| \leq |x_{\hat{n}}-L|+|L-x_{\hat{n}+1}|<1$. This is a contradiction as clearly $|x_{\hat{n}}-x_{\hat{n}+1}|=2$. Therefore the sequence does not converge.
A: We have
$$ x_{2n} = (-1)^{2n} = 1 \to 1,$$
and
$$ x_{2n+1} = (-1)^{2n+1} = (-1)(-1)^{2n} = -1 \to -1.$$
Therefore, $x_n$ is divergent.
