Ideal sampling using sinc funcion
Your professor is using sloppy notation. When one writes the sample value $x[\cdot]$, it is generally accepted that the argument is an integer (which $nT$ might not be) and we relate this to the continuous-time signal $x(t)$ via $x[n] = x(nT)$. So, what your professor should have said is that $y[n]$ has value $T\cdot x[n] = T\cdot x(nT)$, and not that $y[n] = T\cdot x[nT]$ (even if $T$ happens to be an integer).  An outline of a proof that $y[n] = T\cdot x[n] = T\cdot x(nT)$ is given below.

If $x(t)$ is a continuous-time signal bandlimited to $\left(-\frac{1}{2T}, \frac{1}{2T}\right)$, that is, its Fourier transform $X(f)$ has the property that $X(f) = 0$ for all $|f| \geq \frac{1}{2T}$,then it can be reconstructed from its samples spaced $T$ seconds apart. So, let's consider the signal $x(t)\operatorname{sinc}\left(\frac{t}{T}\right)$ where
$$\operatorname{sinc}(t) = \begin{cases}\frac{\sin(\pi t)}{\pi t},&amp; t \neq 0,\\1, &amp; t = 0.\end{cases}$$
The Fourier transform of $\operatorname{sinc}(t)$ is $\operatorname{rect}(f)$ and so the Fourier transform of $\operatorname{sinc}\left(\frac{t}{T}\right)$ is $T\operatorname{rect}(fT)$, and the Fourier transform of $x(t)\operatorname{sinc}\left(\frac{t}{T}\right)$ is the convolution  $X(f)\circledast T\operatorname{rect}(fT)$ of the Fourier transforms $X(f)$ and $T\operatorname{rect}(fT)$ of $x(t)$ and $\operatorname{sinc}\left(\frac{t}{T}\right)$ respectively.  Now, in principle, the convolution of a function and a rectangular pulse is easy to compute but I refuse to write out the answer because it is not needed: what we need to find is the value of $\int_{-\infty}^\infty x(t)\operatorname{sinc}\left(\frac{t}{T}\right) \mathrm dt$ which, if you think about it a bit, is actually the value of the Fourier transform of $x(t)\operatorname{sinc}\left(\frac{t}{T}\right)$ evaluated at $f=0$. That is,
\begin{align}
\int_{-\infty}^\infty x(t)\operatorname{sinc}\left(\frac{t}{T}\right) \mathrm dt &amp;= X(f)\circledast T\operatorname{rect}(fT)\bigg |_{f=0}\\
&amp;= T\int_{-\infty}^\infty X(\lambda)\operatorname{rect}((0-\lambda) T) \, \mathrm d\lambda\\
&amp;= T\int_{-\infty}^\infty X(\lambda)\operatorname{rect}(\lambda T) \, \mathrm d\lambda &amp; \scriptstyle\text{rect is an even function}\\
&amp;= T\int_{-\frac{1}{2T}}^{\frac{1}{2T}}X(\lambda) \, \mathrm d\lambda &amp;\scriptstyle\text{rect has finite support}\\
&amp;= T\int_{-\infty}^\infty X(\lambda) \, \mathrm d\lambda &amp;\scriptstyle{X(\lambda)=0 ~\text{for}~|\lambda| &gt; \frac{1}{2T}}\\
&amp;= T
\cdot x(0).
\end{align}
Thus, $y[0] \overset{\text{def}}{=} \int_{-\infty}^\infty x(t)\operatorname{sinc}\left(\frac{t}{T}\right) \mathrm dt$ equals $T\cdot x(0)$. I leave it as an exercise for the reader to work out the Fourier transform of $\operatorname{sinc}\left(\frac{t-nT}{T}\right)$ and show that $y[n] \overset{\text{def}}{=} \int_{-\infty}^\infty x(t)\operatorname{sinc}\left(\frac{t-nT}{T}\right) \mathrm dt$ works out to be $T\cdot x(nT)$ and so if we define the samples of $x(t)$ as $x[n] = x(nT)$, then we have that $y[n] = T\cdot x[n]$ for all $n$.