Showing the sequence converges $a_{1}=\frac{1}{2}$, $a_{n+1}=\frac{1}{2+a_{n}}$
Indeed, this sequence is not monotonic.
It seems that for $$n = 2k, k \in \mathbb{N}, a_n \geq \sqrt{2} - 1$$ and for $$n = 2k+1, k \in \mathbb{N}, a_n \leq \sqrt{2}-1$$ (this can be proved using induction).
Now, $$a_{2n+2} - a_{2n} = \displaystyle \frac{1}{2+a_{2n+1}} - a_{2n} = \frac{1}{2+\frac{1}{2+a_{2n}}} - a_{2n} = \frac{-2a_{2n}^2-4a_{2n}+2}{5+2a_{2n}}.$$
Because $a_{2n} \geq \sqrt{2}-1 \implies -2a_{2n}^2-4a_{2n}+2 \leq 0$ so $(a_{2n})_{n \in \mathbb{N}}$ is decreasing.
You can prove analogously that $(a_{2n+1})_{n \in \mathbb{N}}$ is increasing.
Now you write the recursion relation for $a_{2n}$ and $a_{2n+1}$, apply $\displaystyle \lim_{n \to \infty}$, and you get that the limit is $\sqrt{2}-1$.
You can also just apply $\displaystyle \lim_{n \to \infty}$ in the starting recursion relation, because $(a_{2n})_{n \in \mathbb{N}}$ decreases to $\sqrt{2} - 1$ and $(a_{2n+1})_{n \in \mathbb{N}}$ increases towards $\sqrt{2}-1$, but writing it for $n$ even and $n$ odd is more rigurous.
$$ a_{2n+2} = \displaystyle \frac{1}{a_{2n+1}+2} = \frac{1}{2+\frac{1}{2+a_{2n}}} .$$ Applying $\displaystyle \lim_{n \to \infty}$, we get that $l = \displaystyle \frac{1}{2+\frac{1}{2+l}} \iff 2l^2 + 4l - 2 = 0$, and the equation has the solutions $l_1 = -1-\sqrt{2}$ and $l_2 = -1+\sqrt{2}$, but since $a_n >0,\forall n \in \mathbb{N}$, the limit is $l = -1+\sqrt{2}$. This is the same for odd $n$.
The approach from https://math.stackexchange.com/a/124288/42969 works here as well:
We have $$ | a_{n+2} - a_{n+1}| = \frac{|a_{n+1}-a_n|}{(2+a_{n+1})(2+a_n)} \le \frac 14 |a_{n+1}-a_n| $$ which implies that $(a_n)$ is a Cauchy sequence and therefore convergent.
And using the approach from https://math.stackexchange.com/a/133878/42969 one can obtain an explicit formula for $a_n$:
Let $u = \sqrt 2 -1 > 0$ and $v = -\sqrt 2 - 1 < 0$ be the solutions of $x^2 + 2x - 1 = 0$. Then an elementary calculation gives $$ \frac{a_{n+1} - u}{a_{n+1}-v} = \frac uv \frac{a_n - u}{a_n-v} $$ and therefore $$ \frac{a_{n} - u}{a_{n}-v} = \left(\frac uv\right)^{n-1} \frac{a_1 - u}{a_1-v} $$ The quotient $u/v$ is negative and its absolute value is less than one. It follows that $a_n \to u$ (and also that the differences $a_n - u$ are alternating positive and negative).
Another look at the problem from the contraction mapping point of view. We will look at the function $f(x)=\frac{1}{2+x}$ s.t. $a_{n+1}=f(a_n)$.
- $\forall x \in \left[0,\frac{1}{2}\right] => f(x) \in \left[0,\frac{1}{2}\right]$. Indeed $$0\leq x\leq \frac{1}{2} \Rightarrow 2\leq 2+x\leq 2+\frac{1}{2} \Rightarrow \frac{1}{2}\geq \frac{1}{2+x}\geq \frac{1}{2+\frac{1}{2}}>0$$ or $$\frac{1}{2}\geq f(x)\geq 0$$
- $f(x)$ is a contraction mapping on $\left[0,\frac{1}{2}\right]$ (it's wider, but we don't need more than this), from MVT $\forall x,y \in \left[0,\frac{1}{2}\right], \exists c$ in between them s.t. $$|f(x)-f(y)|=|f'(c)||x-y|=\left|-\frac{1}{(2+c)^2}\right||x-y|\leq\frac{1}{4}|x-y|$$
Since $a_1 \in \left[0,\frac{1}{2}\right]$, from Banach fixed-point theorem, the sequence has a limit on $\left[0,\frac{1}{2}\right]$ which you can find from $L=\frac{1}{2+L}$ (which, I believe, is what you did).