E.g. when having 2 signals $x(t)$ and $y(t)$ and transforming each one with adding their spectra, the operation is linear, as the result would be the same as the transformation of $(x + y)(t)$.
Even when looking at the sampling series, each of the infinite elements adds up linearly.
But how is aliasing a non-linear operation and how is it proven mathematically?
I think you mean "images", not "aliases". They become aliases if there is foldover from resampling.
It's because you are not adding two signals, $x(t)$ and $\operatorname{III}(t)$, you are multiplying them that these images appear.
$$\begin{align} x_\text{s}(t) & \triangleq x(t) \cdot \operatorname{III}(t/T) \\ &= x(t) \cdot \sum\limits_{n=-\infty}^{+\infty} T \delta(t-nT) \\ &= T \sum\limits_{n=-\infty}^{+\infty} x(t) \cdot \delta(t-nT) \\ &= T \sum\limits_{n=-\infty}^{+\infty} x(nT) \cdot \delta(t-nT) \\ &= T \sum\limits_{n=-\infty}^{+\infty} x[n] \cdot \delta(t-nT) \\ \end{align}$$
where $x[n] \triangleq x(nT)$ and
$$ \operatorname{III}(u) \triangleq \sum\limits_{n=-\infty}^{+\infty} \delta(u-n) $$
$$\begin{align} \operatorname{III}(t/T) &= \sum\limits_{n=-\infty}^{+\infty} \delta\left(\tfrac{t}{T}-n\right) \\ &= \sum\limits_{n=-\infty}^{+\infty} \delta\left(\tfrac{t-nT}{T}\right) \\ &= \sum\limits_{n=-\infty}^{+\infty} T \delta(t-nT) \\ &= \sum\limits_{k=-\infty}^{+\infty} e^{j k \frac{2 \pi}{T} t} \\ \end{align}$$
The last line is from doing fourier series. Now, if you use the shifting property of the Fourier Transform, then the Fourier Transform of $x_\text{s}(t)$ is
$$\begin{align} X_\text{s}(f) & \triangleq \mathscr{F}\{x_\text{s}(t)\} \\ &= \mathscr{F}\{x(t) \operatorname{III}(t/T) \} \\ &= \mathscr{F}\left\{x(t) \sum\limits_{k=-\infty}^{+\infty} e^{j k \frac{2 \pi}{T} t} \right\} \\ &= \mathscr{F}\left\{\sum\limits_{k=-\infty}^{+\infty} x(t) \, e^{j k \frac{2 \pi}{T} t} \right\} \\ &= \sum\limits_{k=-\infty}^{+\infty} \mathscr{F}\left\{ x(t) \, e^{j k \frac{2 \pi}{T} t} \right\} \\ &= \sum\limits_{k=-\infty}^{+\infty} X\left(f-\tfrac{k}{T}\right) \\ \end{align}$$
where $ X(f) \triangleq \mathscr{F}\{ x(t) \} $ .
That non-linear process of multiplication generates frequency components than did not previously exist in $x(t)$. Those new components are simply shifted versions of $X(f)$ and are called "images".
Hints:
The answers to these questions are straightforward and, combined, they answer the original question.
