The squeeze theorem needs the two bounds to share their limit

Given: The unique sequences $(s_k)$ of reals with $s_0=1$ and $s_{\sigma (k)}=-s_k$ for every $k$ (The recursion theorem), and the constant sequences $a_k:=-1$, $b_k:=1$, $x_k:=s_k$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

The sequence $(s_k)$ is the alternating sequence of FALSE: every bounded sequence converges: it is introduced there by the recursion $s_0=1$, $s_{\sigma (k)}=-s_k$ (FALSE: every bounded sequence converges).

$(s_k)$ does not converge (FALSE: every bounded sequence converges).

Absolute value is even: $\mathrm{\mid }-u\mathrm{\mid }=\mathrm{\mid }u\mathrm{\mid }$ (Basic properties of the absolute value).

Induction principle: if $P(0)$ holds and $P(k)$ implies $P(\sigma (k))$ for every $k$, then $P(k)$ holds for every natural $k$ (The principle of mathematical induction).

Absolute value: $\mathrm{\mid }u\mathrm{\mid }=u$ if $u\ge 0$ and $\mathrm{\mid }u\mathrm{\mid }=-u$ if $u<0$; so $\mathrm{\mid }0\mathrm{\mid }=0$, $\mathrm{\mid }1\mathrm{\mid }=1$ and, for any $u$, the equality $\mathrm{\mid }u\mathrm{\mid }=1$ forces $u=1$ when $u\ge 0$ and $u=-1$ when $u<0$ (Absolute value in an ordered field).

Trichotomy: for all reals $u,v$ exactly one of $u<v$, $u=v$, $v<u$ holds, and the order is transitive (Ordered field).

$0<1$, so by adding constants $-1<1$ and in particular $1\mathrm{\ne }-1$, while $-1\le 1$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities).

Convergence: $(z_k)$ converges to $z$ when for every rational $\epsilon >0$ there is $K$ with $\mathrm{\mid }{z}_k-z\mathrm{\mid }<\epsilon$ for all $k\ge K$ (Limits and Cauchy sequences of reals).

Squeeze theorem: if $a_k\le x_k\le b_k$ eventually and $(a_k)$, $(b_k)$ both converge to the same real $L$, then $(x_k)$ converges to $L$ (The squeeze theorem).