Proof by induction concerning approximation algorithm for subset sum
You haven't described the approximation algorithm, so I assume it's something like the following one (perhaps with a few changes in some details):


Input: a universe $U = \{x_1,\ldots,x_n\}$, a target sum $T$, and an accuracy parameter $\epsilon$.
Initialize $L_i = \{0\}$.
For $i = 1, \ldots, n$:

Compute $L_{i+1} = L_i \cup \{\ell + x_i : \ell \in L_i\}$.


Remove all element larger than $T$ from $L_{i+1}$.


Replace each element $\ell \in L_{i+1}$ by the largest power of $1+\epsilon/2n$ which is at most $\ell$.

Return the element in $L_n$ closest to $T$.


This is a $1+O(\epsilon)$-approximation algorithm, in the sense that it can tell whether there is a subset of $U$ summing to roughly $T$, up to a factor of $1+O(\epsilon)$.

The exercise asks you to show that for each $i \in \{0,\ldots,n\}$, the following is true: if $y \in P_i$ (the set of all sums of subsets of $\{x_1,\ldots,x_i\}$) is at most $T$, then there is an element $z \in L_i$ which is approximately equal to $y$, in the sense that
$$ \frac{y}{(1+\frac{\epsilon}{2n})^i} \leq z \leq y. $$
Taking $i = n$ will give the desired analysis of the complete algorithm.

Let us denote the claim symbolically by $C_i$. The exercise suggests that this claim can be proved by induction on $i$. This suggests the following plan to prove $C_0,\ldots,C_n$:


Prove $C_0$.
For $i=1,\ldots,n$, assuming $C_{i-1}$ is true, prove that $C_i$ is true.


To find out what claim $C_0$ or $C_{i-1}$ is, you will need to substitute $0$ or $i-1$ for $i$ in the claim.

Good luck with executing this plan.