The answer is all $c \leq 1$. It is enough to provide a construction for $c=1$. Indeed, consider numbers of the form $\frac{a}{2024^b}$ where $b\geq 0$, $1 \leq a \leq 2024^b$, and $2024 \nmid a$, lexicographically increasing in the order of $(b,a)$. Taking this sequence as $x_n$ works. Conversely, given $c>1$ and a sequence $x_n \in [0,1]$, we have to find a pair $m>n>0$ of integers such that $|x_m-x_n|<\frac{c}{f(m)}$. Assume no such pair exists. Let $N=2024^d$ for $d$ large enough, and say the numbers $x_1,x_2, \dots, x_N$ have been placed. Clearly distance between any two of them is at least $\frac{c}{N}$. Draw open balls of radius $\frac{c}{2024N}$ around each $x_i$. They don't intersect, because $2024>2$. Thus the uncovered area in $[0,1]$ is split into at most $N+1$ closed intervals; let their lengths be $l_1,l_2, \dots, l_{N+1}$ from left to right ($l_1$ or $l_{N+1}$ might be $0$ depending on the leftmost or rightmost point). Now say we have to place $x_{N+1}, \dots, x_{2024N}$. These cannot go into the open balls, and distance between any two of them is at least $\frac{c}{2024N}$. A closed interval of length $l_i$ can accommodate at most $\frac{2024Nl_i}{c}+1$ of them, so in total we can accommodate at most $$N+1+\frac{2024N}{c}\sum_{i=1}^{N+1} l_i$$of them. Now, at least $N-2$ complete open balls are in $[0,1]$, and at the two endpoints, at least one half each are in $[0,1]$. These take up at least $(2(N-2)+2) \cdot \frac{c}{2024N}$ space, so $\sum l_i \leq 1-\frac{(2N-2)c}{2024N}$. Also, there are $2023N$ numbers we need to accomodate. Putting it all together, we get $$2023N \leq N+1+\frac{2024N}{c} \left(1-\frac{(2N-2)c}{2024N} \right)=\frac{2024N}{c}+3-N.$$$$\implies 1 \leq \frac{1}{c}+\frac{3}{2024N}.$$Now just take $N \to \infty$ to get $c \leq 1$.

Ok so the answer is $c \leq 1$.The construction for $c=1$ is to consider the sequence $\frac{a}{2024^b}$ where $2024 \not | a$ and consider the sequence lexicographically.(Like for $b=1,a=1 \cdots 2024$ we define $\{x_i\}_{1\leq i \leq 2024}$ for $b=2,a=1 \cdots $ we define the others (ofc $2024 \not | a$) and so on. The bound is to be obtained like this ,for $1\leq i,j \leq 2024^k$.We have $|x_i-x_j| \geq \frac{c}{2024^k}$.This is true for any consecutive two elements of the sequence (in increasing order).Now we have in total $2024^k-1$ spaces.Adding all these spaces we have the total space as $\frac{c}{2024^k}\cdot(2024^k-1)$.This must be less than $1$.Taking $K \rightarrow \infty$ we get that $c \leq 1$ .The end $\blacksquare$