Finding an $x \in \mathbb R$ so that $\|I - x A\|_2$ becomes minimal

Just minimise the square:

\begin{eqnarray} \|I-xA\|^2 &=& \sum_i \sum_j (\delta_{ij} -x \delta_{ij} A_{ij} + x^2 A_{ij}^2) \\ &=& n - x \sum_i A_{ii} + x^2 \sum_i \sum_j A_{ij}^2 \\ &=& n -x \operatorname{tr} A + x^2 \|A\|_F^2 \end{eqnarray}

Assuming that $A \neq 0$ we have (after differentiating with respect to $x$ and setting to zero): $x = { \operatorname{tr} A \over 2\|A\|_F^2 }$.

The norm can be express with the trace: $$ \forall B \in \mathcal{M}_n(\mathbb{R}) \quad ||B||_2^2 = \text{tr}(B^TB) $$ Using this: $$ ||I - xA||_2^2 = \text{tr}((I - xA)^T(I - xA)) $$ After expanding this you will find a 2nd degree polynomial in $x$. You know how to find the minimum of this.