How can one express $\sqrt{2+\sqrt{2}}$ without using the square root of a square root?

Short answer: you can't. Call your element $\alpha$. You want $\alpha$ to live in a field extension of the form $\mathbb{Q}(\beta^{1/4})$ for some $\beta \in \mathbb{Q}$. It's clear that $L = \mathbb{Q}(\alpha)$ is a degree 4 field extension of $\mathbb{Q}$ (you've written down a minimal polynomial).

One can check easily that $L$ is Galois over $\mathbb{Q}$ (i.e. you can check that $\sqrt{2 - \sqrt{2}} \in L$). But $\mathbb{Q}(\beta^{1/4})$ (for $\beta$ squarefree) is not.

For completeness's sake here is the solution to the original problem:

We first solve the equation $$ \sqrt{2+\sqrt{x}}=x $$ and the only real root $y$ is about 1.812. Now we have $\sqrt{2+\sqrt{2}}\approx 1.847$. If we differentiate $\frac{\sqrt{2+\sqrt{x}}}{x}$ with respect to $x$, the derivative is always negative. So when $s_{n}\ge y$, $s_{n}$'s value would decrease under iteration until it is less than $y$, and when $s_{n}\le y$ it would increase, since on the interval $[0,y]$ the quotient is greater than $0$. This force the limit to be $y$.

Another way to see it is observing that $$ s_{n}=\sqrt{2+\sqrt{s_{n-1}}}, |s_{n+1}-s_{n}|=|\sqrt{2+\sqrt{s_{n}}}-\sqrt{2+\sqrt{s_{n-1}}}|=|\frac{\sqrt{s_{n}}-\sqrt{s_{n-1}}}{s_{n}+s_{n+1}}| $$ therefore we have $$ |\frac{s_{n+1}-s_{n}}{s_{n}-s_{n-1}}|=\frac{1}{|s_{n}+s_{n-1}||\sqrt{s_{n}}+\sqrt{s_{n-1}}|}\le \frac{1}{4} $$ and the rest follows from contraction mapping theorem.