Why is this space dense in this Sobolev space? (Bochner spaces) Let $V$ be a separable Banach space, and let $H$ be a Hilbert space. We have the Hilbert triple
$$V \subset H \subset V^*.$$
By separability, there exists subspaces $V_k$ with $V_k \subset V_{k+1}$ such that $\bigcup_k V_k \subset V$ dense.
Let $W^{1} = \{ u \in L^2(0,T;V) : u' \in L^2(0,T;V^*), u(0) = u(T)\}$ and endow it with the norm
$$\lVert {u}\rVert_{W^1} = \lVert{u}\rVert_{L^2(0,T;V)} + \lVert{u'}\rVert_{L^2(0,T;V^*)}.$$
Let $S_k = \text{span}(\varphi v : v \in V_k, \varphi \in C([0,T]), \varphi(0) = \varphi(T)\}$.
The claim is that $\bigcup_k S_k$ is dense in $W^1.$
No idea why this is true?
 A: First we note that $V_k$ can be choosen to be a finite dimensional space. Now we can find a sequence $v_k$, such that $v_1,...,v_n$ is a basis for $V_n$ for all $n$. Moreover, we choose $v_k$ in such a way that $$\tag{1}\|v_k\|\leq\frac{1}{2^k}$$
Note that $C^1=\{u\in C([0,1];V):\ u'\in C([0,1];V)\ \mbox{and}\ u(0)=u(T)\}$ is dense in $W^1$. This can be showed by using convolution, similar to what is done in Lemma 7.2. of the book you have cited.
Now, the only thinkg we need to prove is that $\cup S_k$ is dense in $C^1$. To this end, fix some $t\in [0,1]$ and note that for this $t$, there exist a family $(\alpha_k(t))$ such that $$\sum_{k=1}^n\alpha_k(t)v_k\to u'(t)\ \mbox{in}\ V\tag{2}$$
Choose $\epsilon>0$ and take $\delta>0$ such that $$\|u'(t)-u'(s)\|\leq\epsilon,\ \mbox{for}\ |t-s|\leq \delta\tag{3}$$
Take $s$ as in $(3)$ and define $\alpha_k(s)=\alpha_k(t)+s-t$. Note that (by using $(2)$ and $(3)$) $$\|\sum_{k=1}^n\alpha_k(s)-u'(s)\|\leq 3\epsilon \tag{4}$$ 
Now, we cover $[0,1]$ with a finite number of open sets, and we construct functions $\alpha_k(s)$ similar to that in $(4)$ to conclude that for each $t$ the convergence in $(2)$ is valid where $alpha_k$ is a continuous function. Because $[0,1]$ is compact, we conclude that the functions $\alpha_k$ are uniformly continuous which implies that the convergen in $(2)$ does not depend on $t$. Therefore, $$u'(t)=\sum_{k=1}^\infty \alpha_k(t)v_k\tag{5}\ \mbox{uniformly}$$
We can therefore integrate $(5)$ to get that $$u(t)=\sum_{k=1}^\infty \int\alpha_k(t)v_k,\ \mbox{uniformly}$$
