I also had troubles with this theorem when I read the Cazenave's book. Recently I found this answer for proving the theorem. However, I think there is still a little flaw in the proof, namely in step 1, we can not directly use that $L^p(I,Y)$ has dual $L^{p'}(I,Y')$, since this is valid for instance under the condition that $Y$ has Radon-Nikodym-property. So I would like to write up my proof and I hope that everything is ok. The basic idea is essentially the same as the one given in the originial proof, but with a minor modification.

WLOG we can assume that $|I|<\infty$, for general case we can show the claim by using monotone convergence theorem. We also restrict ourselves to the case $1<p<\infty$, for $p=\infty$ we can just replace the weak topology arguments by the corresponding weak-$*$-arguments. Since $X$ is reflexive, we know that $X$ has RNP, thus together with the boundedness of $(f_n)_n$ in $L^p(I,X)$ we know that $f_n$ converges weakly to some $h$ in $L^p(I,X)$. Now let $m$ be a number in the interval $(1,\min(p,q))$. From the assumption that $f_n$ is bdd in $L^q(I,Y)$ and the fact that $|I|<\infty$ we know that $f_n$ is bdd in $L^m(I,Y)$, thus using Fatou's lemma one deduces that $f$ is also in $L^m(I,Y)$. Now let $\phi\in L^{m'}(I,Y')$. we want to show that
$$\int_{I}\left<f_n(t)-f(t),\phi(t)\right>_{Y,Y'}dt\to 0 $$
as $n\to\infty$ (**Note I am not showing duality! Since $Y$ does not have RNP**). To see this, since $|I|<\infty$ and the integrand in the above integral converges to zero a.e. (which is shown in the original answer), we can use Egoroff's theorem to decompose $I$ into $K$ and $K^c$ such that the integrand converges to zero uniformly on $K$, while the measure of $K^c$ is small. Using absolute continuity of the integral of an integrable function we can make the integral involving $K^c$ arbitrary small, as long as the measure of $K^c$ is small, which is possible. On the other hand, due to embedding we know that $f_n$ converges to $h$ in $L^m(I,Y)$ weakly, thus we have (notice that $L^{m'}(I,Y')$ is a subspace of $(L^m(I,Y))'$, as long as $Y$ is a Banach space and $m\in[1,\infty)$)
$$\int_{I}\left<f(t)-h(t),\phi(t)\right>_{Y,Y'}dt=0 $$
for all $\phi\in L^{m'}(I,Y')$. Using fundamental lemma of calculus of variations we conclude that $f=h$ a.e., and hence $f$ is in $L^{p}(I,X)$, and $f_n$ converges weakly to $f$ in $L^p(I,X)$. The last statement can now be deduced by the lower semicontinuity of the norm function.