Representing a continuous LTI system as a discrete one
The sequence $h[n]$ consists of samples of a low pass filtered version of $h_c(t)$. Define that low pass filtered impulse response as

$$\tilde{h}_c(t)=\int_{-\infty}^{\infty}h_c(\tau)\frac{\sin(\pi(t-\tau)/T)}{\pi(t-\tau)/T}d\tau\tag{1}$$

and

$$h[n]=\tilde{h}_c(nT)\tag{2}$$

The output signal is

$$\begin{align}y(t)&amp;=(x\star h_c)(t)\\&amp;=\int_{-\infty}^{\infty}x(t-\tau)h_c(\tau)d\tau\\&amp;=\int_{-\infty}^{\infty}\sum_{k=-\infty}^{\infty}x[k]\frac{\sin(\pi(t-kT-\tau)/T)}{\pi(t-kT-\tau)/T}h_c(\tau)d\tau\\&amp;=\sum_{k=-\infty}^{\infty}x[k]\int_{-\infty}^{\infty}h_c(\tau)\frac{\sin(\pi(t-kT-\tau)/T)}{\pi(t-kT-\tau)/T}d\tau\\&amp;=\sum_{k=-\infty}^{\infty}x[k]\tilde{h}_c(t-kT)\tag{3}\end{align}$$

Consequently, we have

$$\begin{align}y(nT)&amp;=\sum_{k=-\infty}^{\infty}x[k]\tilde{h}_c((n-k)T)\\&amp;=\sum_{k=-\infty}^{\infty}x[k]h[n-k]\end{align}\tag{4}$$

This results is also obvious from the fact that $(x\star h_c)(t)=(x\star\tilde{h}_c)(t)$ must hold, because $x(t)$ is band-limited. And since $\tilde{h}_c(t)$ is also band-limited, we can simply take samples from $x(t)$ and $\tilde{h}_c(t)$ to obtain the result.