Unique solution and iteration convergent to the solution

Prove that $x=\sin x+\frac{1}{4}$ has got a unique solution on $[\pi/4,\pi/2]$. Show that the iteration $x_0\in[\pi/4,\pi/2], x_{n+1}=\sin x_n+\frac{1}{4}$ is convergent to that solution.
I tried to show the first part using the graphs of functions $\sin x$ and $x-1/4$. But I am not sure if this is the best way (is it formal...). The next step is a bit harder to me. Should I somehow use the Picard-Lindelof theorem?
Let $$f(x)=\sin x+\frac14.$$ If $\pi/4\le x\le \pi/2$ then $$\frac\pi4<\frac{\sqrt2}{2}+\frac14\le f(x)\le1+\frac14<\frac\pi2.$$ Thus $f([\pi/4,\pi/2])\subset[\pi/4,\pi/2]$. Moreover $$|f'(x)|=|\cos x|\le\cos\frac\pi4=\frac{\sqrt2}{2}<1,\quad x\in[\pi/4,\pi/2].$$ The contraction principle,or Banach fixed point theorem, gives the result.