First of all we need $a_1 \not \in \left (-\dfrac {\sqrt{5} + 1}{2}, \dfrac {\sqrt{5} - 1}{2}\right )$ (the roots of $x^2+x-1=0$). Consider the function $f\colon \mathbb{R} \setminus \left (-\dfrac {\sqrt{5} + 1}{2}, \dfrac {\sqrt{5} - 1}{2}\right ) \to \mathbb{R}_+$ given by $f(x) = \sqrt{x^2+x-1}$. We have $f(x) = x$ if and only if $x=1$, $f(1)=1$, $f(-2)=1$. The graph of $f$ is made by arms of a hyperbola. And $a_{n+1} = f(a_n)$. It is then seen that for $a_1 \in \left (-2, -\dfrac {\sqrt{5} + 1}{2}\right ]\cup \left [\dfrac {\sqrt{5} - 1}{2}, 1\right )$ the "sequence" $(a_n)_{n\geq 2}$ is decreasing, and gets at some point below the value $\dfrac {\sqrt{5} - 1}{2}$, absurd.

Am I missing something or the fact that the sequence decreases does not explain why $ a_n$ cannot go to some limit above $\dfrac{\sqrt{5}-1}{2}$ ?

Because, as $(a_n)_{n\geq 2}$ being decreasing, if it stayed above $\dfrac {\sqrt{5}-1}{2}$, then it would be convergent to a limit $\ell \geq \dfrac {\sqrt{5}-1}{2}$. However, passing to the limit will then yield $\all = 1$, in contradiction with $a_2<1$.

Suppose that $a_1 \in (-2;1)$. We proceed in several steps. Step 1: We have $0 \le a_i <1$ for all $i \ge 2$ Proof: It is rather obvious that all terms (with the index of at least $2$) of the sequence must be positive. Denote $a_1 = x$. Since inequality $1>x^2 +x-1$ holds for all $x \in (-2,1)$, we get that $a_2 <1$. Now note that if $a_i <1$, then $a_{i+1}^2 = a_i^2+a_i -1<1+1 -1 < 1 \implies a_{i+1} <1$. This proves the first step. Step 2: $a_2, a_3, \ldots$ is strictly decreasing sequence. Proof: Indeed, it sufficient to check that $a_{i+1}^2 = a_i^2+a_i -1 < a_i \implies a_i <1$, which obviously holds from the result obtained in the previous step. From before established results, we conclude that we express $a_2, a_3, \ldots, a_n$ in the following form: $$ a_2 = 1 - \epsilon_2, a_3 = 1 - \epsilon_3, \ldots, a_n = 1 - \epsilon_n$$where $0< \epsilon_2 < \epsilon_3 < \ldots < \epsilon_n$ Step 3: Finish. Proof: We have the following identities: \begin{align*} a_3^2 - a_2^2 = a_2- 1 \\ a_4^2 - a_3^2 = a_3 -1 \\ \ldots \\ a_{k+1}^2 - a_k^2 =a_k -1 \end{align*}Summing together yields: $$ a_{k+1}^2 =a_2+a_3 + \ldots+a_k - (k-1) +a_2^2= (1- \epsilon_2)^2 - \sum_{i=2}^{k}\epsilon_i $$But $\sum_{i=2}^{k}\epsilon_i $ can grow arbitrary large; therefore meaning that eventually we will get $(1- \epsilon_2)^2 - \sum_{i=2}^{k}\epsilon_i <0 \implies a_{k+1}^2 < 0$ - contradiction. This solves the problem.

First, we note that $a_i \ge 0$ for $i \ge 2$. We have the following claim: Claim 1: If $a_2 <1$ then $1>a_2>a_3 > \cdots $. Proof: It is enough to show that $a_n < 1$ implies that $a_{n+1} < a_n < 1$. We have $$a_{n+1}^2 = a_n^2+a_n -1 < a_n < 1.$$This finishes the proof of claim. $\spadesuit$ Claim 2: $a_2 \ge 1$. Proof: Suppose $a_2 < 1$. Then by Claim $1$, $a_2, a_3, \cdots$ is a decreasing sequence, and they are bounded below by $0$, thus by Monotone Convergence Theorem they converge at some value $\alpha$. Now, taking limit as $n \to \infty$ on both sides of $a_{n+1}^2 = a_n^2+a_n -1$, we get $$\alpha^2 = \alpha^2 + \alpha -1 \implies \alpha = 1.$$But this is clearly impossible, since $a_i \le a_2$ for all $i \ge 2$ implies $\alpha \le a_2 < 1$. $\spadesuit$ Now, it suffices to note that Claim 2 immediately finishes the problem, since $$a_1^2 + a_1 - 1 = a_2^2 \ge 1 \implies (a_1 +2)(a_1-1) \ge 0. \ \square$$