Let $\epsilon \in (0, 1)$. Consider the following sequence with $a_0 = \sqrt{2\epsilon}$.
$$a_n = a_{n-1} + \sqrt{2(a_{n-1} + \epsilon)}$$
I would like to bound this sequence as $a_n \leq f(n, \epsilon)$. Writing down the first few terms, we have \begin{align} a_0 &= \sqrt{2\epsilon}\\ a_1 &= \sqrt{2\epsilon} + \sqrt{2(\sqrt{2\epsilon} + \epsilon)}\\ a_3 &= \sqrt{2\epsilon}+\sqrt{2(\sqrt{2\epsilon}+\epsilon)}+\sqrt{2(\epsilon+\sqrt{2\epsilon}+\sqrt{2(\sqrt{2\epsilon}+\epsilon)})} \end{align}
My guess is that it should look like $a_n \leq c(n)\epsilon^{1/2^n}$ or something similar to this. Any advice on how to proceed?
As @Alex suggests it is convenient to consider $b_n = a_n+\epsilon$ since you have that $$b_{n+1} - b_n = \sqrt{2 b_n}\,.$$
You have that $b_0 = \epsilon + \sqrt{2\epsilon}$.
Now consider $u(t)$ the solution of the Cauchy problem $$\dot u(t) = \sqrt{2 u(t)}\,;$$ $$u(0) = \epsilon + \sqrt{2\epsilon}\,.$$
You have $$u(t) = \frac{1}{2}\left(t + \sqrt{2}\sqrt{\epsilon+\sqrt{2\epsilon}}\right)^2\,.$$
Now you can show that $a_n\leq u(n)-\epsilon$ for all $n$.
Indeed you have that for each $n$ there exists a $x\in(n,n+1)$ such that $\dot u(x) = u(n+1) - u(n)$. Since $u$ has positive second derivative, $\dot u(x)>\dot u(n)$. By definition of $u$ you have $$u(n+1) - u(n) \geq \dot u(n) = \sqrt{2u(n)}\,.$$
Since $u(0) = b_0$ you have $$u(1) = u(0) + (u(1)-u(0)) \geq u(0) + \sqrt{2u(0)} = b_0+\sqrt{2b_0} = b_1\,.$$ By induction you show that if $u(n)\geq b_n$ then $$u(n+1) = u(n) + (u(n+1)-u(n))\geq u(n) + \sqrt{2u(n)}\geq b_n+\sqrt{2b_n} = b_{n+1}\,.$$
So for all integer $n$ you have $u(n)-\epsilon\geq b_n-\epsilon = a_n$.