For compactness: topologically, you can see $M_2(\mathbb{C})$ as $\mathbb{C}^4$. This means that convergence in the operator norm in $M_2(\mathbb{C})$ is simply convergence in coordinates, and it implies that Heine-Borel holds; so all you need to show is closedness. If you have a Cauchy sequence in $[a\mathbb{I},b\mathbb{I}]$, a limit matrix will exist because in each entry you will have a Cauchy sequence of complex numbers. Moreover, since the characteristic polynomial's entries depend continuously on the entries of the matrix, the characteristic polynomial of the limit will be limit of the characteristic polynomials of the sequence, which shows that the limit will again be in $[a\mathbb{I},b\mathbb{I}]$. So the interval is compact.
Convexity: the interval $[a\mathbb{I},b\mathbb{I}]$ can be characterized as those hermitian matrices $A$ such that $ a\,x^Tx\leq x^TAx\leq b\,x^Tx,\ \ x\in\mathbb{C}^2. $ So, if $A,B\in[a\mathbb{I},b\mathbb{I}]$, $\gamma\in[0,1]$, $ a\,x^Tx=\gamma\,a\,x^Tx+(1-\gamma)a\,x^Tx\leq\gamma\,x^TAx+(1-\gamma)x^TBx=x^T(\gamma A+(1-\gamma)B)x ; $ similarly, we obtain $ x^T(\gamma A+(1-\gamma)B)x \leq b\,x^Tx. $