Ok, this might be an overkill and there is nice solution by Aryabhata, but I wanted to present another approach using Banach fixed-point theorem. I am not 100% sure I didn't do any mistakes, so if somebody could comment on this I would really appreciate it.
To prove the main question let $\Gamma\phi(t) = \sqrt{1+2\int_0^t\phi(s)\,\mathrm{d}s}$. Then $f_0 = f$, $f_{n+1} = \Gamma f_n$ is non-decreasing sequence of functions for which $f_\infty = \lim_{n\to\infty}f_n$ is a fixed point of $\Gamma$, thus $f_\infty(t) = 1 + t$. To prove this, let's bound:
\begin{align*} (\Gamma\phi_1 - \Gamma\phi_2)(t) &\leq^{(*)} \int_0^{t}(\phi_1-\phi_2)(s)\,\mathrm{d}s \\\ &\leq t\|\phi_1 - \phi_2\|_\infty \end{align*}
so
\begin{align*} \|\Gamma^2\phi_1 - \Gamma^2\phi_2\|_\infty &= (\Gamma^2\phi_1 - \Gamma^2\phi_2)(t_\max) \\\ &\leq \int_0^{t_\max} (\Gamma\phi_1 - \Gamma\phi_2)(s)\,\mathrm{d}s \\\ &\leq \int_0^{t_\max} s\|\phi_1-\phi_2\|_\infty\,\mathrm{d}s \\\ &\leq \frac{1}{2}\|\phi_1 - \phi_2\|_\infty \end{align*}
Therefore $\Gamma^2$ is a contraction mapping on $C[0,1]$ with the standard $\|\cdot\|_\infty$ norm so $\Gamma^2$ has a unique fixed-point which happens to be $f_\infty$ (both even and odd indexes converge to the same fixed point). So the limit exists and we have that $f(t) \leq f_\infty(t) = 1 + t$ what completes the proof.
Small technical lemma: $\sqrt{1+x} - \sqrt{1+y} \leq \frac{x-y}{2}\text{ for all }x \geq y \geq 0 \tag{$*$}$ Proof: $\sqrt{1+x}-\frac{x}{2}$ is non-increasing for $x\geq 0$.