The answer is NO without further assumptions on $f$ other than it is integrable. For a counter-example, consider the function
$$f(x) = (-1)^{\left\lfloor x^2 \right\rfloor}$$
For any $k \in \mathbb{Z}_{+}$, define
$$F_k = \int_{\sqrt{2k-2}}^{\sqrt{2k}} f(x) dx = -\sqrt{2k-2} + 2\sqrt{2k-1} -\sqrt{2k}$$
Notice for large $k$,${}^{\color{blue}{[1]}}$
$$\begin{align}F_k
&= (\sqrt{2k-1} - \sqrt{2k-2})-(\sqrt{2k}-\sqrt{2k-1})\\
&= \frac{1}{\sqrt{2k-1}+\sqrt{2k-2}} - \frac{1}{\sqrt{2k}+\sqrt{2k-1}}\\
&= \frac{2}{(\sqrt{2k-1}+\sqrt{2k-2})(\sqrt{2k}+\sqrt{2k-1})(\sqrt{2k}+\sqrt{2k-2})}\\
&\sim \frac{1}{8\sqrt{2k^3}}
\end{align}
$$
We find as a sequence in $n$, $\displaystyle\;\int_0^{\sqrt{2n}} f(x) dx = \sum_{k=1}^n F_k$ converges to some limit $\Delta$ as $n \to \infty$.${}^{\color{blue}{[2]}}$
For any $\epsilon > 0$, choose a $N \in \mathbb{Z}_{+}$ so large such that
$$\sqrt{2N+1}-\sqrt{2N} < \frac{\epsilon}{2} \quad\text{ and }\quad \left|\sum_{k=1}^m F_k - \Delta \right| < \frac{\epsilon}{2},\forall m \ge N$$
For any $y > \sqrt{2N}$, let $n \ge N$ be the integer such that $y \in \left[\sqrt{2n},\sqrt{2n+2}\right]$, we have
$$\left|\int_0^y f(x) dx - \int_0^{\sqrt{2n}} f(x) dx\right| = \left|\int_{\sqrt{2n}}^y f(x) dx\right| \le \sqrt{2n+1}-\sqrt{2n}\\ \le \sqrt{2N+1}-\sqrt{2N} < \frac{\epsilon}{2}$$
This leads to
$$\left| \int_0^y f(x) dx - \Delta \right| \le \left| \int_{\sqrt{2n}}^{y} f(x) dx \right| + \left| \sum_{k=1}^n F_k - \Delta \right| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$
Since $\epsilon$ is arbitrary, $\displaystyle\;\lim\limits_{y\to\infty}\int_0^y f(x)dx\;$ exists and equal to $\displaystyle\;\Delta = \sum_{k=1}^\infty F_k\;$.
By definition, $|f(x)| = 1$ for all $x$. It is impossible to build a sequence $x_n$ such that $f(x_n)$ converges to $0$.
Random Notes
$\color{blue}{[1]}$ - Another way to see the $F_k \sim O\left(\frac{1}{\sqrt{k^3}}\right)$
dependence goes like this. The expression
$$F_k = -\sqrt{2k-2} + 2\sqrt{2k-1} - \sqrt{2k}$$
has the form of $2^{nd}$ order finite difference for function $-\sqrt{x}$ at $x = 2k-1$. By a generalization of mean value theorem to higher order finite differences, there is a $\xi \in (2k-2,2k)$ such that
$$F_k = \left.\frac{d^2}{dx^2}(-\sqrt{x})\right|_{x=\xi} = \frac{1}{4\sqrt{\xi^3}} \sim \frac{1}{8\sqrt{2k^3}}$$
$\color{blue}{[2]}$ - It can be shown $\Delta = 2(1-\sqrt{8})\zeta\left(-\frac12\right) \approx 0.76020962521937$.
To begin, the function $(x,y) \mapsto e^{-xy} \sin x$ is not absolutely integrable. It is enough to show this for the region $[0,\infty)^2$.
If it were then we could apply Fubini's theorem to the iterated integrals and conclude
$$\int_0^\infty \int_0^\infty e^{-xy} | \sin x| \, dx \, dy = \int_0^\infty \int_0^\infty e^{-xy} | \sin x| \, dy \, dx $$
However,
$$\int_0^\infty e^{-xy} | \sin x| \, dy= \frac{|\sin x|}{x} $$
and this is neither Lebesgue nor improperly Riemann integrable over $[0,\infty]$.
The question seems to be is the improper Riemann integral $\int_{\mathbb{R}^2} f$ convergent, where $f(x,y) = e^{-xy} \sin x$.
For integrals over $\mathbb{R}$ the definition of an improper integral as $\lim_{a \to -\infty, b \to +\infty}\int_a^b f$ is unambiguous. For multiple integrals the definition of improper integral requires more care. The standard is
$$\tag{*}\int_{\mathbb{R}^2}f = \lim_{n \to \infty}\int_{A_n}f,$$
where $A_n$ is an admissible sequence, that is a sequence of compact Jordan measurable sets such that $A_n \subset A_{n+1}$ for all $n$ and $\cup_{n=1}^\infty A_n = \mathbb{R}^2$.
For the improper integral to be well defined, (*) must hold for any admissible sequence with convergence to a unique value. This has an important consequence:
If $\int_{\mathbb{R}^2} f$ converges as an improper integral in the
sense of (*) then $\int_{\mathbb{R}^2} |f|$ converges.
See here for a proof.
Since $\int_{\mathbb{R}^2} e^{-xy} |\sin x|$ does not converge it follows that the improper integral $\int_{\mathbb{R}^2} e^{-xy} \sin x$ does not converge in the sense of (*). In fact, it can take on different values depending on exactly how the limiting process is defined. This is analogous to the problem with rearrangements of conditionally convergent infinite series. An example is given here.
Best Answer
$ \sin(x) $ won't do, but it's only a tad trickier. $$ f(x) = \sin\left(\ln(x)\right) $$ should do the job (you can integrate it exactly by elementary techniques and then show it is $ \mathcal{O}(T) $).
Note that the function is bounded and continuous in $ (0,+\infty) $.