We have shown the continuity of the Gamma function for all $x \in (1,\infty)$. Regarding this interval I have understood all the steps.
However, when it comes to prove continuity for $x\in(0,1]$ the professor just wrote that this follows from the general property of the Gamma function: $\Gamma(x+1)=x\Gamma(x)$. I don't see why? Maybe someone can explain it to me.
Since $1/x$ is continuous on $(0,1]$ and $\Gamma(x+1)$ is continuous on $(0,1]$ (because $\Gamma$ is continuous on $(1,2]$ by your previous result), then $$ \Gamma(x) = \frac{1}{x}\,\Gamma(x+1) $$ is continuous as well on $(0,1]$ as a product of two continuous functions ;)