Let $\mathbf X := (X, \mathcal S, \mu, T)$ be an ergodic measure preserving system with finite measure such that for every increasing sequence $\{n_k\}$ of natural numbers with positive lower density, we have that for all $f \in L^1(X)$,
$$\frac{1}{N} \sum_{i=0}^{N-1} f(T^{n_i} x) \to \int_X f d\mu$$
as $n \to \infty$ for a.e. $x \in X$.
Question: Is $\mathbf X$ necessarily weakly mixing?
I think $\mathbf{X}$ has to be weakly mixing.
In order to prove it, note that one characterization of weak mixing is that the Koopman operator $U_Tf:= f\circ T$ does not have any eigenvectors on the space $L^{2}_{0}$ of mean zero square-integrable functions. Suppose then that there is an eigenvector $f$ such that $f\circ T = \lambda f$. Your assumption says that for any increasing sequence $(n_k)$ of positive lower density we have $$ \lim_{N\to \infty} \left(\frac{1}{N} \sum_{k=0}^{N-1} \lambda^{n_k}\right) f(x) = 0 $$ for a.e. $x\in X$. It means that $\lim_{N\to \infty} \frac{1}{N} \sum_{k=0}^{N-1} \lambda^{n_k} = 0$. We now have to find an appropriate sequence $(n_k)$ for which this convergence does not hold.
Note that the sequence $(\lambda^n)$, where $\lambda = \exp(i\varphi)$, is an orbit of the rotation by an angle $\varphi$ on the unit circle. If $\lambda$ is a root of unity of order $m$ then we can take $n_k:= km$ so that $\lambda^{n_k}=1$ -- we have convergence to $1$. If $\lambda$ is not a root of unity then we have an irrational rotation, so the orbit is equidistributed on the unit circle. In particular it means that the set of natural numbers for which $\lambda^n$ lands in a small neighbourhood around $1$ has positive lower density. We can use this subset as our sequence $(n_k)$ and then $\frac{1}{N} \sum_{k=0}^{N-1} \lambda^{n_k}$ cannot converge to $0$.
