Vector-valued integral: Equivalence of Riemann sums and Rudin's definition

Question

Definition/Notation: In his book "Functional Analysis", Rudin defines the integral of a vector-valued function as follows (Definition 3.26).
Let $(Q, \mu)$ be a measure space, $X$ a topological vector space on which the dual $X^*$ separates points, and $f:Q\to X$ such that $\lambda f:Q\to\mathbb{R}$ is integrable w.r.t. $\mu$ for any $\lambda\in X^*$. If there exists some $y\in X$ such that $$ \lambda(y)=\int_Q \lambda f d\mu \qquad (*) $$ for all $\lambda \in X^*$, define $\int_Qfd\mu=y$.
Question: I would like to show that this is equivalent to defining it as the strong limit of Riemann sums in the following way (Chapter 3, Exercise 23):
Suppose $Q$ is a compact Hausdorff space, $\mu$ a Borel probability measure, $X$ a Fréchet space [for my purposes we could even assume Banach, if that makes it easier] and $f:Q\to X$ continuous. (The existence of the integral under these circumstances is shown in Theorem 3.27.) Then for any neighborhood $V\subseteq X$ of $0$, there exists a partition $E_1, ..., E_n$ of $Q$ (i.e. $E_i\cap E_j = \emptyset$ for $i\neq j$ and $\bigcup_{i=1}^n E_i=Q$) such that $$ z:=\int_Q fd\mu - \sum_{i=1}^n \mu(E_i)f(s_i) \in V $$ for any choice of $s_i\in E_i$.
Own work: As mentioned, I assume $(X, \lVert \cdot \rVert_X)$ is Banach. Given some neighborhood $0\in V\subseteq X$, let $B(0,\varepsilon)\subseteq V$ for some $\varepsilon>0$. By the Hahn-Banach theorem, there exists $\lambda \in X^*$ with $\lambda(z)=\lVert z \rVert_X$ and $\lVert \lambda \rVert=1$, so $$ \lVert z \rVert_X = |\lambda(z)| = \left|\int_Q \lambda fd\mu - \sum_{i=1}^n \mu(E_i)\lambda(f(s_i))\right| \le \sum_{i=1}^n \int_{E_i} |\lambda(f(t)-f(s_i))|\mu(dt) \le \varepsilon \qquad (**) $$ given that we define the $E_i$ such that $f(t)-f(s) \in B(0,\varepsilon)$ whenever $t,s \in E_i$. Since $Q$ is compact, $f$ is uniformly continuous, which I assume helps with finding the partition. I struggle, since it has to be a finite partition (which would be easy if $X$ were totally bounded, but it doesn't have to be). Could someone help me?