Prime Subsequence 
Question:
Prove that for any $\alpha > 0 $, there exists a non-decreasing sequence of integers $\mathcal{A}=\{n_1, n_2, \cdots \}$, such that
$$p_{n_j} \sim \alpha j$$ as $j\to \infty$.

I have totally no idea how to start with. The only thing I know is that from prime number theorem, we have $p_n \sim n\log n$. This means that $p_n$ should grow slightly faster than $n$, so I do not understand why the subsequence could grow in the order of $j$, which should be even smaller than $n_j$. Any help would be appreciated.
Edit:
There is a disproof for the case when we only require $n_1<n_2<\cdots$.
Therefore we are now interested in the case when $n_1\leq n_2\leq \cdots$
 A: Here's a disproof based off your intuition:
$$\lim_{j\to\infty}\frac{p_{n_j}}{\alpha j}=\lim_{j\to\infty}\frac{n_j\ln(n_j)}{\alpha j}$$
But we know $n_j\geq j$ which implies
$$\lim_{j\to\infty}\frac{n_j\ln(n_j)}{\alpha j}\geq \lim_{j\to\infty}\frac{j\ln(j)}{\alpha j}=\lim_{j\to\infty}\frac{\ln(j)}{\alpha}=\infty$$

EDIT: I'm including this edit because I believe I have found the source for OP's question. See number $4$ here. In this question
$$n_1\leq n_2\leq n_3\leq ...$$
Basically, we can create a subsequence which repeats primes (a weakly increasing subsequence).

EDIT 2: OP has modified the question to allow
$$n_1\leq n_2\leq n_3\leq ...$$
This will provide an answer to this new question:
We will construct a sequence. For all $j$, let $n_j$ be the index such that
$p_{n_j}$ is the closest prime to $\alpha j$ (note for later that $n_j\to\infty$). If two primes are equidistant from $\alpha j$, choose the smaller of the two. In this manner, we have precisely defined our sequence $p_{n_j}$. Now, define
$$g_n=p_{n+1}-p_n$$
Since $g_n$ defines the distance between consecutive primes and $p_{n_j}$ is the closest prime to $\alpha j$, we have
$$\alpha j-g_{n_j-1}<p_{n_j}<\alpha j+g_{n_j}$$
Define $q_j=n_{j+1}-n_j$ and note that
$$\alpha(j+1)-\alpha (j-1)=2\alpha$$
implies $q_j<3\lceil \alpha \rceil$ (for the sake of notation, call this $M$). This is because there less than $M$ integers between $\alpha(j+1)$ and $\alpha(j-1)$. That is, $n_j$ can jump by at most $M$ each time it increases. Of course, this is a terrible bound and could be greatly improved but it is sufficient for this proof. This bound implies
$$p_{n_j}<p_{jM}$$
Continuing, we have
$$\lim_{j\to\infty} \frac{p_{n_j}}{\alpha j}\leq \lim_{j\to\infty} \frac{\alpha j+g_{n_j}}{\alpha j}=1+\lim_{j\to\infty} \frac{g_{n_j}}{\alpha j}$$
It is known (see here) that for all but a finite number of cases
$$g_n<p_n^\theta$$
where $\theta=\frac{249}{250}<1$. This implies
$$1+\lim_{j\to\infty} \frac{g_{n_j}}{\alpha j}\leq 1+\lim_{j\to\infty} \frac{p_{n_j}^\theta}{\alpha j}\leq 1+\lim_{j\to\infty} \frac{p_{jM}^\theta}{\alpha j}$$
$$=1+\lim_{j\to\infty} \frac{(jM)^\theta(\ln(jM))^\theta}{\alpha j}=1+\frac{M^\theta}{\alpha}\lim_{j\to\infty} \frac{(\ln(jM))^\theta}{j^{1-\theta}}$$
Since $1-\theta=\frac{1}{250}>0$, this limit is equal to $0$ giving us
$$\lim_{j\to\infty} \frac{p_{n_j}}{\alpha j}\leq 1$$
For the lower bound, we have
$$\lim_{j\to\infty} \frac{p_{n_j}}{\alpha j}\geq \lim_{j\to\infty} \frac{\alpha j-g_{n_j-1}}{\alpha j}=1- \lim_{j\to\infty} \frac{g_{n_j-1}}{\alpha j}\geq 1- \lim_{j\to\infty} \frac{p_{n_j-1}^\theta}{\alpha j}\geq 1- \lim_{j\to\infty} \frac{p_{n_j}^\theta}{\alpha j}$$
From here, we analysis proceeds in the same manner to conclude
$$1\leq \lim_{j\to\infty} \frac{p_{n_j}}{\alpha j}$$
We conclude
$$\lim_{j\to\infty} \frac{p_{n_j}}{\alpha j}=1$$
