= Solution
Let $F(x)=\sqrt{2+\sqrt x}$ for $x\geq0$. This is an increasing <continuous function>. Starting at $a_1=\sqrt2$, induction gives $\sqrt2\leq a_n\leq2$: if $a_n\leq2$, then $a_{n+1}\leq\sqrt{2+\sqrt2}<2$, and every iterate is at least $\sqrt2$.
The first step has $a_2>\sqrt2=a_1$. Since $F$ is increasing, $a_n\geq a_{n-1}$ implies $a_{n+1}=F(a_n)\geq F(a_{n-1})=a_n$. Thus this <sequence> is increasing and bounded above. Let $L_-=\sup_n a_n$. For every $\varepsilon>0$, a term exceeds $L_--\varepsilon$; all later terms lie between it and $L_-$. This proves convergence, the elementary <bounded monotone sequence theorem>.
For the second starting value, write the iterates as $b_1=4$ and $b_{n+1}=F(b_n)$. Here $b_2=2<4$. Monotonicity of $F$ propagates this inequality, so $b_n$ is decreasing. It is bounded below by $\sqrt2$ because every positive iterate is at least that value. Hence it converges to its <infimum> $L_+$. From $b_2=2$ onward all terms lie in $[\sqrt2,2]$.
<Continuity> of $F$ permits passing to each <limit> in the recurrence, giving $L_\pm^2=2+\sqrt{L_\pm}$. To prove that these <fixed points> agree, suppose $r,s\geq\sqrt2$ both satisfy that equality. Subtracting the two equations and rationalizing gives
$$
0=(r-s)\left(r+s-\frac1{\sqrt r+\sqrt s}\right).
$$
The second factor is positive: $r+s\geq2\sqrt2$, whereas $(\sqrt r+\sqrt s)^{-1}\leq1/(2\sqrt{\sqrt2})<2\sqrt2$. Thus $r=s$. Therefore
$$
\boxed{L_-=L_+=L,\qquad \sqrt2<L<2,\qquad L^2=2+\sqrt L.}
$$
Existence has been supplied by the two convergent <sequences>, and the subtraction argument gives the required uniqueness in their common interval.
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