The condition (b) tells that if $x \neq y$ are in $X$, then at least one of the four numbers
$z_1 = x - (y - x), \quad z_2 = x + \frac{1}{3}(y - x), \quad z_3 = x + \frac{2}{3}(y - x), \quad z_4 = x + 2(y - x)$,

is in $X$.

Assume that $X$ is finite and let $x_0$ be the smallest element of $X$. By condition (a), the set $X \backslash \{x_0\}$ is nonempty. Consider a map
$$f: X \backslash \{x_0\} \rightarrow X \backslash \{x_0\}$$
which satisfies, for $y \in X \backslash \{x_0\}$, the condition
$$f(y) \in \{x_0 + \frac{1}{3}(y - x_0), x_0 + \frac{2}{3}(y - x_0), x_0 + 2(y - x_0)\} \cap X .$$
Notice that $f(y)$ is defined for all $y \in X \backslash \{x_0\}$ by condition (b) since $x_0 - (y - x_0) < x_0$ is not in $X$.

Now, fix an element $y_0 \in X \backslash \{x_0\}$ and define by induction $a_0 = y_0 - x_0$ and $a_{n+1} = f(x_0 + a_n) - x_0$ for all $n \in \mathbb{N}_0$. Clearly, the sequence ($a_n$) is well defined and positive and we have, for each $n \in \mathbb{N}_0$, either
$$a_{n+1} = \frac{1}{3} a_n, \quad \text{or} \quad a_{n+1} = \frac{2}{3} a_n, \quad \text{or} \quad a_{n+1} = 2 a_n .$$
Therefore, for each positive integers $m, n$, there exist non-negative integers $p, q$ with $p + q > 0$ such that
$$a_{m+n} = \frac{2^p}{3^q} a_m .$$
Because $X \backslash \{x_0\}$ is finite, there exist two positive integers $m_0, n_0$ such that $a_{m_0 + n_0} = a_{m_0}$. Hence the corresponding two non-negative integers $p_0, q_0$ with $p_0 + q_0 > 0$ satisfy
$$1 = \frac{2^{p_0}}{3^{q_0}},$$
which is impossible.

This proves that $X$ contains infinitely many elements.