Suppose that $b_n^{2} \rightarrow4$. Prove that $b_n$ has a subsequence that converges to $2$ or $-2$.
I'm not really sure on the approach, I see that simply taking the root yields something similar, but doesn't say much about a subsequence and doesn't really seem like a proof.
On
Since $(b_n^2)$ is convergent, $(b_n^2)$ is bounded, hence $(b_n)$ is bounded. Bolzano-Weierstraß gives: there is a subsequence $(b_{n_k})$ which is convergent. Let us denote the limit of this subsequence by $b$.
Then $ b_{n_k}^2 \to b^2$ as $k \to \infty.$
But $ (b_{n_k}^2)$ is a subsequence of $(b_n^2)$, hence $ b_{n_k}^2 \to 4$ as $k \to \infty.$
This gives $b^2=4.$
On
Define
Let $P_1$ be the condition that a subsequence of $b_n$ converges to $-2$.
$$\forall N_1,\epsilon_1\gt0,\exists n\ge N_1:|b_n+2|\lt\epsilon_1$$
Let $P_2$ be the condition that a subsequence of $b_n$ converges to $2$. $$\forall N_2,\epsilon_2\gt0,\exists n\ge N_2:|b_n-2|\lt\epsilon_2$$
Assume
Assume that $\lnot(P_1\lor P_2)$, which is equivalent to $\lnot P_1\land\lnot P_2$.
$\lnot P_1$: no subsequence of $b_n$ converges to $-2$. $$\exists N_1,\epsilon_1\gt0:\forall n\ge N_1,|b_n+2|\ge\epsilon_1$$
$\lnot P_2$: no subsequence of $b_n$ converges to $2$. $$\exists N_2,\epsilon_2\gt0:\forall n\ge N_2,|b_n-2|\ge\epsilon_2$$
Since $\lnot P_1\land\lnot P_2$, we get that $$\forall n\ge\max(N_1,N_2),\left|b_n^2-4\right|\ge\epsilon_1\epsilon_2$$
However, this means that $b_n^2$ does not converge to $4$, which is a contradiction.
Conclude
Therefore, we must have $P_1\lor P_2$.
If $b_n \ge 0$ then $$ |b_n^2 -4| = |b_n+2| \cdot |b_n-2| \ge 2 |b_n-2| \implies |b_n-2| \le \frac 12 |b_n^2 -4| $$ and if $b_n < 0$ then $$ |b_n^2 -4| = |b_n-2| \cdot |b_n+2| \ge 2 |b_n+2|\implies |b_n+2| \le \frac 12 |b_n^2 -4| \, . $$ At least one of these cases must occur infinitely often, i.e. there is a subsequence $(b_{n_k})$ of non-negative elements (which converges to $+2$) or there is a subsequence $(b_{n_k})$ of negative elements (which converges to $-2$).