Prove that if the linear system $Ax = b$ is perturbed in both $A$ and $b$
$$(A + \delta A)(x+ \delta x) = b + \delta b$$
then
$$ \frac{ \| \delta x \| }{ \|x\| } \leq \frac{\kappa(A)}{1-\kappa(A)\frac{ \|\delta A\| }{ \|A\| }} \left(\frac{ \| \delta A \| }{ \|A\| } + \frac{ \| \delta b \| }{ \|b\| } \right)$$
proof (Attempt)
First note that in the inequality above that the denominator term on the right hand side must obey
$$0< 1-\kappa(A)\frac{ \|\delta A\| }{ \|A\| } < 1 \tag{1}$$
because norms, and therefore condition numbers are always positive. The upper bound of $1$ is achieved by the fact that $\|\delta A\| << \|A\|$. And therefore the denominator factor in question, in $(1)$, has multiplicative inverse greater than $1$, implying
$$ \kappa(A) \left(\frac{ \| \delta A \| }{ \|A\| } + \frac{ \| \delta b \| }{ \|b\| } \right)< \frac{\kappa(A)}{1-\kappa(A)\frac{ \|\delta A\| }{ \|A\| }} \left(\frac{ \| \delta A \| }{ \|A\| } + \frac{ \| \delta b \| }{ \|b\| } \right)$$
So for the proof all we must show is
$$ \frac{ \| \delta x \| }{ \|x\| } \leq \kappa(A) \left(\frac{ \| \delta A \| }{ \|A\| } + \frac{ \| \delta b \| }{ \|b\| } \right)$$
From the perturbed equation \begin{align} (A + \delta A)(x+ \delta x) &= b + \delta b \\ Ax +A\delta x + \delta Ax + \delta A\delta x &= b + \delta b \\ A\delta x &= \delta b - \delta Ax \end{align}
The double infinitesimal term is dropped (I don't know why, my book does this quite a bit, how is this justified?). Multiplying the pseudoinverse $A^+$, taking the norm, dividing by the norm of $x$, and using the triangle inequality, we get
\begin{align} \frac{\|\delta x \|}{\|x\|} \leq \|A^+\|\left(\frac{\|\delta b \|}{\|x\|} + \frac{\|\delta Ax \|}{\|x\|} \right) \end{align}
Note that from $b = Ax$ we have $\|b\| \leq \|A\|\|x\|$. Using that fact
\begin{align} \frac{\|\delta x \|}{\|x\|} \leq \|A^+\|\left(\frac{\|\delta b \|}{\|x\|} + \frac{\|\delta Ax \|}{\|x\|} \right) &\leq \|A^+\|\left(\frac{\|\delta b \|}{\|b\|}\|A\| + \frac{\|\delta Ax \|}{\|x\|}\right) \\ \\ &= \|A^+\|\left(\frac{\|\delta b \|}{\|b\|}\|A\| + \|\delta A \|\right) \\ \\ &= \kappa(A)\left(\frac{\|\delta b \|}{\|b\|} + \frac{|\delta A \|}{\|A\|}\right) _\square \end{align}
The justification for dropping that double infinitesimal is bothering me.
This proof does not make much sense for me because you are trying to prove the inequality $$\tag{1} \frac{\|\delta x\|}{\|x\|}\leq\frac{\kappa(A)}{1-\kappa(A)\frac{\|\delta A\|}{\|A\|}}\left(\frac{\|\delta A\|}{\|A\|}+\frac{\|\delta b\|}{\|b\|}\right) $$ which follows from $$\tag{2} (A+\delta A)(x+\delta x)=b+\delta b $$ without dropping the second order term $\delta A\delta x$ by proving an inequality which might not be true because you dropped the second order term (2).
You do not need to drop anything to prove (1), just need to do a bit more manipulations. See, e.g., my answer here.