To prove your proposition, we need to get some relatively accurate estimates of $jx\bmod 1$ for $1 \le j \le N$. One option to consider is a consequence of Dirichlet's approximation theorem that
$$\left\lvert x - \frac{p}{q} \right\rvert \lt \frac{1}{q^2} \tag{1}\label{eq1A}$$
is satisfied by infinitely many integers $p \ge 0$ and $q \gt 0$. This works if $\gcd(p, q) = 1$, but we can't guarantee that. Nonetheless, by ensuring $\frac{1}{q^2}$ is small enough (and, so, $q$ large enough), we can effectively get that by having $\frac{p}{q}$, in lowest terms, using a large enough denominator. To do this, for any positive integer $n$, have
$$m = \min\left(\left.\left\lbrace\left\lvert x - \frac{r}{s} \right\rvert\;\;\right|\; \; s\in\mathbb{Z}, \; 1 \le s \le n, r\in\mathbb{Z},\; 0 \le r \le s \right\rbrace\right) \tag{2}\label{eq2A}$$
Choose any $p$ and $q$ which satisfies \eqref{eq1A}, as well as that $\frac{1}{q^2} \lt m$. Let
$$d = \gcd(p, q), \;\; p = dp_1, \;\; q = dq_1, \;\; \gcd(p_1, q_1) = 1 \tag{3}\label{eq3A}$$
Substituting this into \eqref{eq1A} gives
$$\left\lvert x - \frac{dp_1}{dq_1} \right\rvert \lt \frac{1}{(dq_1)^2} \;\;\to\;\; \left\lvert x - \frac{p_1}{q_1} \right\rvert \lt \frac{1}{(dq_1)^2} \le \frac{1}{q_1^2} \tag{4}\label{eq4A}$$
From \eqref{eq2A} and that $\frac{1}{q^2} \lt m$, we get that $q_1 \gt n$. Let $N = q_1$.
First, consider where $x - \frac{p_1}{q_1} \gt 0$. We then get for all integers $1 \le j \le N$ that
$$0 \lt x - \frac{p_1}{N} \lt \frac{1}{N^2} \;\;\to\;\; \frac{jp_1}{N} \lt jx \lt \frac{jp_1}{N} + \frac{j}{N^2} \tag{5}\label{eq5A}$$
Note that with $k = jp_1 \bmod N$, i.e., $0 \le k \lt N$, then since $0 \lt \frac{j}{N^2} \le \frac{1}{N}$, we have $jx \bmod 1 \in I_k$. If there are $2$ values of $j$, e.g., $j_1$ and $j_2$ with $j_2 \gt j_1$, where $j_{1}p_{1}$ has the same remainder as $j_{2}p_{1}$ when divided by $N$, we would get
$$N \mid j_{2}p_{1} - j_{1}p_{1} \;\;\to\;\; N \mid p_1(j_2 - j_1) \tag{6}\label{eq6A}$$
With $N = q_1$ and \eqref{eq3A}'s condition of $\gcd(p_1, N) = 1$, then also
$$N \mid j_2 - j_1 \tag{7}\label{eq7A}$$
However, $0 \lt j_2 - j_1 \lt N$, so this is not possible. This means each $jx \bmod 1$ is in a separate interval of length $\frac{1}{N}$ of $[0, 1]$.
The other case where $x - \frac{p_1}{q_1} \lt 0$ can be handled similarly to show that it also satisfies your proposition. Thus, in either case, your proposition is true.
