Proving $\text {limsup}_{x\to \infty}x\int_{x}^{x+1}\sin (t^2) \mathrm{d}t=1$ It is not difficult through the substitution $u=t^2$ to show that $$x\int_{x}^{x+1}\sin (t^2) \mathrm{d}t=\frac 12 \left(\cos(x^2)-\cos[(x+1)^2]\right)+r(x)$$
where $\lim_{x \to \infty} r(x) =0$, it seems intuitive that $\frac 12 \left(\cos(x^2)-\cos[(x+1)^2]\right)$ should always be able to reach $1$ no matter how far down along the $x$ axis we are. The solutions manual says that $\frac 12 \left(\cos(x^2)-\cos[(x+1)^2]\right)=\sin(s^2+\frac 14)\sin(s)$ where $s=x+\frac 12$ and then gives the following explanation

"Choose any integer $n >\frac {2-\epsilon}{8\epsilon}$ then the interval $I=\left(\frac 14 + \left(\left(2n +\frac 12 \right)\pi -\epsilon \right)^2,\frac 14 + \left(\left(2n +\frac 12 \right)\pi +\epsilon \right)^2\right)$ is longer than $2\pi$ and thus there is a $t\in I$ such that $\sin(t^2+\frac 14)=1$. But then $tf(t)>1-\epsilon$ (where $f(t)=\int_{t}^{t+1}\sin (u^2) \mathrm{d}u$) it follows that $\text {limsup}_{x\to \infty}x\int_{x}^{x+1}\sin (t^2) \mathrm{d}t=1$."

I understand why, given that $tf(t)>1-\epsilon$ for all $\epsilon$ would imply the statement, but what I don't get is why did the manual forget about the $\sin (t)$ that had come in the simplification. It is possible that $\sin (t)<1-\epsilon$ and then the argument does not work. I also understand the strategy of the proof, building a sequence of $t$'s that converge to $1$, but I cannot figure out where did the bounds for the interval $I$ come from. Is there no simpler, even more general strategy for proving this (or related) proposition?

Your initial idea of enforcing the substitution $t\to \sqrt{t}$ is a good one.  Indeed we have
$$\int_x^{x+1} \sin(t^2)\,dt=\int_{x^2}^{(x+1)^2}\frac{\sin(t)}{2\sqrt{t}}\,dt$$
Now, let's integrate by parts with $u=\frac{1}{2\sqrt{t}}$ and $v=-\cos(t)$. Then, we have
$$\begin{align}
\int_x^{x+1} \sin(t^2)\,dt&=\frac{\cos(x^2)}{2x}-\frac{\cos((x+1)^2)}{2(x+1)}-\frac14 \int_{x^2}^{(x+1)^2}\frac{\cos(t)}{t^{3/2}}\,dt\\\\
&=\frac{1}{2x}\left(\cos(x^2)-\cos((x+1)^2)\right)+\underbrace{\frac{\cos((x+1)^2)}{x(x+1)}-\frac14 \int_{x^2}^{(x+1)^2}\frac{\cos(t)}{t^{3/2}}\,dt}_{=O\left(\frac1{x^2}\right)}\\\\
&=\frac1x \sin\left(\frac{(x+1)^2+x^2}{2}\right)\sin\left(\frac{(x+1)^2-x^2}{2}\right)+O\left(\frac1{x^2}\right)\\\\
&=\frac1x \sin\left((x+1/2)^2+1/4\right)\sin\left(x+1/2\right)+O\left(\frac1{x^2}\right)\\\\
\end{align}$$


The term $\displaystyle \sin(x+1/2)$

Now, let $1>\epsilon>0$ be given.  Suppose that $x+1/2=(2n+1/2)\pi+\nu$, where $\nu\in [-\epsilon,\epsilon]$.  
Then, $\sin(x+1/2)=\cos(\nu)$ and $\cos(\epsilon)\le \sin(x+1/2)\le 1$.  


The term $\displaystyle \sin((x+1/2)^2+1/4)$

We now examine the argument $(x+1/2)^2+1/4$.  
Note that for $x\in [(2n+1/2)\pi-\epsilon,(2n+1/2)\pi+\epsilon]$, the argument $(x+1/2)^2+1/4$ satisfies the inequalities  
$$((2n+1/2)\pi -\epsilon)^2+1/4\le (x+1/2)^2+1/4\le ((2n+1/2)\pi -\epsilon)^2+1/4$$

If we select $n$ large enough such that the length of the interval $[((2n+1/2)\pi -\epsilon)^2+1/4,((2n+1/2)\pi +\epsilon)^2+1/4]$ is at least $2\pi$, then we can select $x\in [(2n+1/2)\pi - \epsilon,(2n+1/2)\pi+\epsilon]$ such that $\sin((x+1/2)^2+1/4)=1$.  
Hence, we require that 
$$((2n+1/2)\pi+\epsilon)^2+1/4-((2n+1/2)\pi-\epsilon)^2-1/4\ge 2\pi$$
This inequality is satisfied if $n\ge \frac{1-\epsilon}{4\epsilon}$.  In that case, there exists an $x$ such that $\sin((x+1/2)^2+1/4)=1$.

Finally, for such an $x$ we have
$$x\int_x^{x+1}\sin(t^2)\,dt=\sin(x+1/2)+O\left(\frac1x\right)$$
Recall that $\cos(\nu)\le \sin(x+1/2)\le 1$, where $\nu\in[-\epsilon,\epsilon]$.  Since $\epsilon>0$ is arbitrary, then we have
$$\limsup_{x\to \infty}x\int_x^{x+1}\sin(t^2)\,dt=1$$
as was to be shown!
