If $f^2(t) \le 1+2\int_0^tf(s)\mathrm{d}s$ prove $f(t)\le 1+t$ If $f(x)$ is positive and continuous on $[0,1]$ and $f^2(t) \le 1+2\int_0^tf(s)\mathrm{d}s$, prove that $f(t)\le 1+t$.
Here's my thinking.
$$f^2(t) \le 1+2\int_0^tf(s)\mathrm{d}s \Rightarrow f(t)\le \sqrt{1+2\int_0^tf(s)\mathrm{d}s} $$ $$\Rightarrow \int_0^t\frac{f(t)}{\sqrt{1+2\int_0^tf(s)\mathrm{d}s}}\mathrm{d}t = \sqrt{1+2\int_0^tf(s)\mathrm{d}s}-1 \le t$$ $$ \Rightarrow 2\int_0^tf(s)\mathrm{d}s\le t^2+2t$$
This comes from the integration of two sides of $f(t)\le 1+t$. But I don't know what to do next.

You are almost done. You correctly derived that
$$
\sqrt{1+2\int_0^tf(s)\mathrm{d}s} - 1 \le t \, .
$$
Now use the initial given inequality to conclude that
$$
f^2(t) \le 1+2\int_0^tf(s)\mathrm{d}s \le (1+t)^2
\implies f(t) \le 1+t \, .
$$
