Here is a clarified version of your proof. (In fact, it was a part of my original solution, but I modified it by following your idea.)
Step 1. Estimation of $xf(x)$
Let $t = \sqrt{u}$. Then
\begin{align*} xf(x) &= x\int_{x}^{x+1} \sin (t^2) \, dt = x \int_{x^2}^{(x+1)^2} \frac{\sin u}{2\sqrt{u}} \, du \\ &= x \left[ \frac{1-\cos u}{2\sqrt{u}} \right]_{x^2}^{(x+1)^2} + x \int_{x^2}^{(x+1)^2} \frac{1-\cos u}{4u^{3/2}} \, du \end{align*}
Since
\begin{align*} x \left[ \frac{1-\cos u}{2\sqrt{u}} \right]_{x^2}^{(x+1)^2} &= \frac{x}{2} \left(\frac{1-\cos\big((x+1)^2\big)}{x+1} - \frac{1 - \cos\big(x^2\big)}{x} \right) \\ &= \frac{1}{2} \left(\cos \left( x^2 \right) - \cos\left((x+1)^2\right) \right) - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)} \\ &= \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right) - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)}, \end{align*}
we have
$xf(x) = \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right) - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)} + x \int_{x^2}^{(x+1)^2} \frac{1-\cos u}{4u^{3/2}} \, du. $
It is easy to observe that
$ - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)} + x \int_{x^2}^{(x+1)^2} \frac{1-\cos u}{4u^{3/2}} \, du = O\left(\frac{1}{x}\right). $
Indeed, it follows from the estimation
\begin{align*} \left| x \int_{x^2}^{(x+1)^2} \frac{1-\cos u}{4u^{3/2}} \, du \right| &\leq x \int_{x^2}^{(x+1)^2} \frac{\left| 1-\cos u \right| }{4u^{3/2}} \, du \\ &\leq x \int_{x^2}^{(x+1)^2} \frac{1}{2x^3} \, du \leq \frac{2x+1}{2x^2} = O\left(\frac{1}{x}\right). \end{align*}
Step 2. Evaluation of limit superior and limit inferior
The estimation above in particular implies that
$ \limsup_{x\to\infty} xf(x) = \limsup_{x\to\infty} \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right)$
and likewise for the liminf. To find the limsup, note from the identity above that
$\limsup_{x\to\infty} xf(x) \leq 1. $
We claim that it is indeed 1. To this end, it suffices to find a subsequence $(x_n)$ such that $x_n \to \infty$ and
$ \sin \left(x_n +\frac{1}{2}\right)\sin\left(x_n^2+x_n+\frac{1}{2}\right) \to 1. $
Let $x = \sqrt{2\pi k}$. Then we have
$ \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right) = \sin^2 \left(\sqrt{2\pi k}+\frac{1}{2}\right) $
As $\sqrt{2\pi k} \to \infty$ and $\sqrt{2\pi(k+1)} - \sqrt{2\pi k} \to 0$ as $k \to \infty$, we can find a subsequence $(k_n)$ such that
$ \min \left\{ \left| \sqrt{2\pi k_n} + \frac{1}{2} - \left( m+\frac{1}{2}\right)\pi \right| : m \in \Bbb{Z} \right\} \to 0 \quad \text{as} \quad n \to \infty. \tag{1}$
Thus for $x_n = \sqrt{2\pi k_n}$ we obtain $ x_n f(x_n) \to 1$ as $n \to \infty$ and therefore $\limsup_{x\to\infty} xf(x) = 1$. A slight modification of this argument also proves that $\liminf_{x\to\infty} xf(x) = -1$.
Proof of the claim $(1)$
Let $\epsilon > 0$. Then there exists $N$ such that whenever $k \geq N$ we have $0 < r_k < \epsilon$, where $r_k$ denotes $r_k = \sqrt{2\pi(k+1)} - \sqrt{2\pi k}$ for simplicity. Then for the sequence of open balls
$ B_k = \left(\sqrt{2\pi k} - r_k, \sqrt{2\pi k} + r_k\right) = \left(\sqrt{2\pi k} - r_k, \sqrt{2\pi (k+1)}\right), $
we have $B_k \cap B_{k+1} \neq \varnothing $ and hence
$ \bigcup_{k=N}^{\infty} B_k = \left(\sqrt{2\pi N} - r_{N}, \infty \right). $
Here, we used the fact that $\sqrt{2\pi k} \to \infty$ as $k \to \infty$. Then for any sufficiently large integer $m$ we have
$ \left(m+\frac{1}{2}\right)\pi - \frac{1}{2} \in \left(\sqrt{2\pi N} - r_{N}, \infty \right). $
Thus we can pick some $k \geq N$ satisfing
$ \left(m+\frac{1}{2}\right)\pi - \frac{1}{2} \in B_k \subset \left(\sqrt{2\pi N} - \epsilon, \sqrt{2\pi N} + \epsilon \right). $
This proves the following proposition: For any $\epsilon > 0$ there exists a positive integer $k$ such that
$ \min \left\{ \left| \sqrt{2\pi k} + \frac{1}{2} - \left( m+\frac{1}{2}\right)\pi \right| : m \in \Bbb{Z} \right\} < \epsilon.$
Then the claim $(1)$ immediately follows.