Let $(g_{k})_{k≥1}$ be the sequence of primes gaps (https://en.wikipedia.org/wiki/Prime_gap#Lower_bounds). I am asking about the possibility of finding a real sequence $(x_{k})_{k≥1}$ with the following properties:
(1) $0<x_{k}<1$
(2) $g_{k}<1/x_{k}$ for all $k>0$. Here I understand that the inequality holds true for all finite primes.
(3) The sequence $x_{k}$ does not have a limit when $k$ goes to infinity.
If condition (2) leads to a contradiction, then we can replace it by the statement:
(2)': $g_{k}<1/x_{k}$ for infinitely many indices $k>0$.
With your updated question, the answer is yes, you can find a real sequence with those $3$ properties. You can choose any relatively small $\epsilon \gt 0$ and then set $x_k = \frac{1}{g_k + \epsilon}$. This obviously meets condition (1), as well as condition (2) as it's equivalent to $x_k \lt \frac{1}{g_k}$.
As for condition (3), note there is no upper limit on $g_k$ since, for example, for any $n \gt 1$, you have that $n! + 2$ to $n! + n$ are all composite. However, recently in $2013$, Yitang Zhang proved, e.g., as explained in Bounded Gaps Between Primes, there's an integer $\lt 7 \times 10^7$ for which there are an infinite number of prime gaps. This was later improved to $246$ (e.g., as stated in the last paragraph of the Upper bounds section of Wikipedia's "Prime Gaps" article).
This combination of no upper limit on the prime gaps and that there are infinitely many prime gaps of less than or equal to $246$ shows the sequence of $(x_{k})_{k≥1}$ will jump between near $0$ and a value $\ge \frac{1}{246 + \epsilon}$ infinitely often and, thus, it does not have a limit.