Convergence in distribution implies convergence in expectation with respect to a continuous, bounded function I am having some trouble understanding the underlined part in the solution here:

This is my thinking: as n goes to infinity, we can write the Yn in the 'last expression' as Y due to (a). Then this expression can be upper-bounded with:
$\Sigma_{i=2}^k (||f||+\epsilon)[F_Y(c_i)-F_Y(c_{i-1}] + 2\epsilon ||f|| $
$\leq (||f||+\epsilon)(F_Y(c_k)-F_Y(c_1))+ 2\epsilon ||f|| \leq (||f||+\epsilon)(1-2\epsilon)+2\epsilon ||f|| = ||f||+\epsilon -2\epsilon^2$ . So I need some help to bound this expression close to $E(f\circ Y)$ as the solution instructs.
Any help is appreciated!

First, as you said, you let $n\to \infty$ and by a) $F_{Y_n}\to F_{Y}$ (I suppose that you're definition of convergence in distribution is pointwise convergence of $F_n$).
Hence you got:
$$ \sum_i (f(c_i)+\eta) [F_Y(c_i) - F_Y (c_{i-1}) + \tilde{\varepsilon} .$$
This $\tilde{\varepsilon}$ is the error term; forget about it. The first piece, up to an (other) error term which I called $\eta$ ( only to distinguish it from the $\varepsilon$), is equal to $\mathbb{E}[f\circ Y] $: you can convince yourself recalling the definition of expected value and integral as a Riemann sum: it's a limit of a sum. Here we do not have the limit, we have the error term $\eta$. You do not have to find any bound, nor write it explicitly, you know that it's almost your expected value.
Than you repeat the procedure with $-f$.
