eigenvalues of a matrix $A$ plus $cI$ for some constant $c$

I'd order the proof correctly: $Ax=\lambda x$ and $cIx = cIx$, so $Ax + cIx =\lambda x +cIx$. Then rearrange to get $(A+cI)x=(\lambda+c)x$.

Note that if $Av = \lambda v$, then $(A+cI)v = (\lambda+c)v$.

Also, $\chi_{A+cI}(x) = \det (xI-A-cI) = \det( (x-c)I -A) = \chi_A(x-c)$.