A useful criterion in vector integration
Here is an answer when $p<\infty$ and $I$ bounded. The argument might be adapted to $p=\infty$ by replacing some weak $L^p$ convergences below by the weak-* $L^{\infty}$ one, and I really believe $I$ bounded is not an issue (otherwise argue locally on any $J\subset I$)
Just a few obvious remarks first: if $X$ is reflexive and $f_n$ bounded in $L^p(I,X)$ with $p<\infty$ (resp. $p=\infty$) then by the Banach-Alaoglu theorem we can extract a subsequence $f_{n_k}\rightharpoonup \tilde{f}$ in this topology (resp. $f_{n_k}\overset{*}{\rightharpoonup} \tilde{f}$). Also, $X\hookrightarrow Y$ implies $ Y'\hookrightarrow X'$.
Claim: $f_n\rightharpoonup f$ in $L^p(I,Y)$. To see this, fix a test function $\varphi\in L^{p'}(I,Y')$ and let $$ h_n(t):=\left<f_n(t),\varphi(t)\right>_{Y,Y'}. $$ By the OP's assumptions we have $$ h_n(t)\to h(t)=<f(t),\varphi(t)>_{Y,Y'}\quad \text{a.e. }t\in I, $$ but also since $f_n\in L^{p}(I,X)\Rightarrow f_n(t)\in X$ a.e. and $Y'\hookrightarrow X'$ $$ h_n(t)=\left<f_n(t),\varphi(t)\right>_{X,X'}\in L^1(I). $$ My claim would immediately follow from $h_n\to h$ in $L^1(I)$, which we can get by Vitali's convergence theorem. In order to apply the latter we only need to check: (i) that $|I|<\infty$ (OK since I'm considering bounded $I$), (ii) $h_n(t)\to h(t)$ for a.e. $t\in I$ (OK by the OP's assumption), (iii) that $|h(t)|<\infty$ for a.e. $t$ (still OK because $f_n(t)\rightharpoonup f(t)$ in $Y$ and the test function is fixed so $\varphi(t)\in Y'$ for a.e. $t$), and finally (iv) that $\{h_n\}_{n}$ is uniformly integrable. The only delicate point is to check (iv), which by the Dunford-Pettis is equivalent to showing that $\{f_n\}_{n}$ is relatively compact for the weak $\sigma(L^1,L^{\infty})$ topology. In order to check this relative compactness, recall from my preliminary remark that $f_{n_k}\rightharpoonup \tilde{f}$ in $L^{p}(I,X)$. For any fixed $\lambda\in L^{\infty}(I)$ we have $\lambda\varphi\in L^{p'}(I,Y')\hookrightarrow L^{p'}(I,X')$ so along this subsequence \begin{align*} \int_I h_{n_k}(t)\lambda(t) & =\int_I\left<f_{n_k}(t),\varphi(t)\right>_{X,X'}\lambda(t)\\ & =\int_I\left<f_{n_k}(t),\lambda\varphi(t)\right>_{X,X'}\\ &\underset{k\to\infty}{\to} \int_I\left<\tilde{f}(t),\lambda\varphi(t)\right>_{X,X'}\\ & =\int_I \tilde{h}(t)\lambda(t) \end{align*} with $\tilde{h}(t)=\left<\tilde{f}(t),\varphi(t)\right>_{X,X'}$. This means that $\tilde{h}\in L^{1}(I)$ is a cluster point for the $\sigma(L^1,L^{\infty})$ topology. Thus $\{h_n\}_{n}$ is weakly $L^1$ relatively compact hence also uniformly integrable, and my claim follows.
Improvement to $L^p(I,X)$. From step 1 and my preliminary remark we have now $$ f_n\rightharpoonup f\text{ in }L^{p}(I,Y),\qquad f_{n_k}\rightharpoonup \tilde{f}\text{ in }L^{p}(I,X). $$ Note that the whole sequence converges for the weak $Y$ norm, but only up to a subsequence for the stronger $X$ norm. Restricting to test-functions $\varphi\in L^{p'}(I,Y')\subset L^{p'}(I,X')$ and by uniqueness of the limit it follows that $f=\tilde{f}$ in $L^{p}(I,Y)$. In particular $f\in L^{p}(I,X)$, and because $\tilde{f}$ was obtained extracting a weakly $L^p(I,X)$ converging subsequence we get $$ ||f||_{L^p(I,X)}=||\tilde{f}||_{L^p(I,X)}\leq \liminf\limits_{n\to\infty} ||f_n||_{L^p(I,X)} $$ as desired and the proof is complete.
Note that I never used the $L^q(I,Y)$ bound, so this assumption can probably be removed. The pointwise convergence in $Y$-weak a.e. $t$ is a quite strong assumption to start with, so this is not surprising.