Questions in my solution about the evaluation of a limit of a recursive sequence

[Ozma]

Let $(a_n)_n$ be a sequence such that for all $n \in \mathbb{N}$ it is $0<a_n<1$ and $a_n(1-a_{n+1}) >\frac{1}{4}$.
Prove that $(a_n)_n$ converges and find its limit.

I've tried this: since $0<a_n<1$ for all $n \in \mathbb{N}$ it is $a_n(1-a_{n+1}) >\frac{1}{4} \iff a_n > \frac{1}{4(1-a_{n+1})}$, and it is $a_{n+1} \ge \frac{1}{4(1-a_{n+1})} \iff (2a_{n+1}-1)^2 \le 0 \iff n \in \mathbb{N}$. So it is $a_{n+1} \leq \frac{1}{4(1-a_{n+1})} < a_n \implies a_{n+1} < a_n$ and so the sequence is decreasing.
Since by hypothesis the sequence is bounded below, it has a real limit $l$: from $a_n(1-a_{n+1})>\frac{1}{4}$ taking the limit in the inequality if follows that it is
$$l(1-l) \ge \frac{1}{4} \iff \left(l-\frac{1}{2}\right)^2 \leq 0 \iff l=\frac{1}{2}$$I am not sure about this because I'm not sure if my proof of the decreasing sequence is correct (I think it is because I've proved that $a_{n+1}$ is less than something that is less than $a_n$, could this work?) and I'm not sure when I take the limit in the inequality, because I have an estimation of the possible limit $l$ and not an equality (but I think that this works because the estimation leads to the fact that $l$ can only be $\frac{1}{2}$ or the other true sentences I've written wouldn't hold; is this correct?). Thank you.