This answer is a response to a comment by the OP on on yoda's answer.
Suppose that $h(t)$, the impulse response of a continuous-time
linear time-invariant system, has the property that
$$\int_{-\infty}^{\infty} |h(t)| \mathrm dt = M$$ for
some finite number $M$. Then, for each and every
bounded input $x(t)$, the output $y(t)$ is bounded also.
If $|x(t)| \leq \hat{M}$ for all $t$ where $\hat{M}$
is some finite number, then $|y(t)| \leq \hat{M}M$ for all $t$
where $\hat{M}M$ is also a finite number.
The proof is straightforward.
$$\begin{align*}
|y(t)| &= \left |\int_{-\infty}^\infty h(\tau)x(t - \tau)\mathrm d\tau\right |\\
&\leq \int_{-\infty}^\infty |h(\tau)x(t - \tau)|\mathrm d\tau\\
&\leq \int_{-\infty}^\infty |h(\tau)|\cdot|x(t - \tau)|\mathrm d\tau\\
&\leq \hat{M}\int_{-\infty}^\infty |h(\tau)|\mathrm d\tau\\
&= \hat{M}M.
\end{align*}$$
In other words, $y(t)$ is bounded whenever $x(t)$ is bounded.
Thus, the condition
$\displaystyle\int_{-\infty}^{\infty} |h(t)| \mathrm dt < \infty$
is sufficient for BIBO-stability.
The condition $\displaystyle\int_{-\infty}^{\infty} |h(t)| \mathrm dt < \infty$
is also necessary for BIBO-stability.
Assume that every bounded input
produces a bounded output. Now consider the input
$x(t) = \text{sgn}(h(-t)) ~\forall~ t$. This is clearly bounded,
($|x(t)| \leq 1$ for all $t$), and at $t=0$, it produces output
$$\begin{align*}
y(0) &= \int_{-\infty}^\infty h(0-\tau)x(-\tau)\mathrm d\tau\\
&= \int_{-\infty}^\infty h(-\tau)\text{sgn}(h(-\tau))\mathrm d\tau
&= \int_{-\infty}^\infty |h(-\tau)|\mathrm d\tau\\
&= \int_{-\infty}^\infty |h(t)|\mathrm dt.
\end{align*}$$
Our assumption that the system is BIBO stable means that $y(0)$ is
necessarily finite, that is,
$$\int_{-\infty}^{\infty} |h(t)| \mathrm dt < \infty$$
The proof for discrete-time systems is similar with the obvious
change that all the integrals are replaced by sums.
Ideal LPFs are not BIBO-stable
systems because the impulse response is not absolutely integrable,
as stated in the answer by yoda. But his answer does not really answer the question
Can anyone give me a proof that ideal LPF can indeed be BIBO unstable?
A specific example of a bounded input signal that produces an unbounded output
from an ideal LPF (and thus proves that the system is not BIBO-stable)
can be constructed as outlined above (see also my comment on the main question).
