Is the set $\{\sqrt n - \lfloor\sqrt n\rfloor : n\in \mathbb Z^+\}$ dense in $[0,1]$?

I was reviewing basic Analysis and thought of this question:

Is the set $\{\sqrt n - \lfloor\sqrt n\rfloor : n\in \mathbb Z^+\}$ dense in $[0,1]$?

Lest we have any ambiguity with notational stuff, $\lfloor \cdot \rfloor$ denotes the integer part of a real number i.e., for $x\in\mathbb R$, $\lfloor x\rfloor := \max\{n\in\mathbb Z\mid n\leq x\}$.

The function $f:\mathbb Z^+\to[0,1]$ defined by $f(n)=\sqrt n - \lfloor\sqrt n\rfloor$ is injective when it is restricted to the prime numbers. I showed this using contradiction.

Let $p_1$ and $p_2$ be distinct prime numbers such that $f(p_1)=f(p_2)$. It would mean that $\sqrt {p_1} + \sqrt{p_2}$ is an integer. (contradiction!)

Thus, the set in our discussion has infinitely many points.

Suppose $p_0$, $p_1$, $\ldots$ is an enumeration of prime numbers. Using pigeon-hole principle, for every $n\geq 1$, there exists primes $p_i$ and $p_j$ with $0\leq i<j\leq n$ such that $$ |f(p_i)-f(p_j)|< \frac{1}{n}$$

I am not sure, how to proceed after this.