In the solution, it was stated that "$x(t)$ periodic implies $y(t)$ is periodic" So I wondered if it's always the case that if $x(t)$ is periodic then so is the output from the convolution, I tried doing research about this in hope of finding any supporting evidence to this but I couldn't, and "circular convolution" is all around the place.
So I wondered if it's always the case that if x(t) is periodic then so is the output from the convolution
Yes, it's always the case. Convolution is a linear operation and linear operations preserve periodicity.
Roughly speaking: any periodic signal consists of a sum of discrete sine waves. The output of an LTI system to a sine wave is also a sine wave at the same frequency. Hence the LTI system preserves the "discreteness" of the original input.
I am somehow rephrasinge what @Hilmar already answered. In the continuous-time setting, convolution cannot invent frequencies. Indeed, there is an intimate relationship between Fourier and convolution, that is not always taught properly. Recently, Michael Bronstein was Deriving convolution from first principles
Have you ever wondered what is so special about convolution? In this post, I derive the convolution from first principles and show that it naturally emerges from translational symmetry.
The basic assumptions that a system is linear, and invariant by shift or in time (LTI) imply the concept of convolution. And convolution implies that a sine undergoing an LTI system will still be a sine, albeit with a time-shift (phase change) and amplitude modification (it can even vanish). In other symbols, through an LTI:
$$ \sin(\omega t)\to A\sin(\omega t+\phi)$$
$A$ can be zero, so frequencies can disappear. Yet if $\sin(\omega t)$ is not present in the original data (no $\omega$), it cannot reappear. Frequency creation from nothing is a feature of nonlinear or time-variant systems.