Showing the sequence $\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$ tends to $2$ using the Epsilon-Neighbourhood definition

Let's make an observation: for each natural number $n$, your sequence can be given explicitly by $$a_n = 2^{\sum_{i=1}^n (\frac{1}{2})^i}$$

For example, $a_2 = \sqrt{2\sqrt{2}} = (2(2^{1/2}))^{1/2} = 2^{1/2} \cdot 2^{1/4} = 2^{1/2 + 1/4}$

Furthermore, we know that $${\sum_{i=1}^n \left(\frac{1}{2}\right)^i} \to 1 \ \ \text{as} \ \ n \to \infty$$

We also know that the function $f(x) = 2^x$ is a continuous function. In other words, if we pick $\epsilon > 0$, there is $\delta > 0$ such that $|x-1| < \delta$ implies $|f(x) - f(1)|< \epsilon$.

Thus, because $$\sum_{i=1}^n \left(\frac{1}{2}\right)^i \to 1$$there is $M > 0$ so that $n > M$ implies $$\left|{\sum_{i=1}^n \left(\frac{1}{2}\right)^i -1}\right| < \delta$$

Hence, letting $x = \sum_{i=1}^n \left(\frac{1}{2}\right)^i$ in the paragraph where I mentioned continuity, we obtain $$\left|f\left(\sum_{i=1}^n \left(\frac{1}{2}\right)^i\right) - 2\right| < \epsilon$$

This is precisely $|a_n -2 | < \epsilon$