Relaxation of the knapsack constraints
Let me call $
\mathcal{X}_1 = \{x \in \mathcal{Z}^n: w^{\top}x \leq b \}
$
and
$
\mathcal{X}_2 = \{x \in \mathcal{Z}^n: \lfloor w\rfloor ^{\top}x \leq \lfloor b\rfloor \}
$ to avoid confusion.

Let $x \in \mathcal{X}_1$. Then: $$b \ge w^T x = \sum_{i : x_i = 1} w_i$$ which implies:
 $$\lfloor b \rfloor \ge  \left\lfloor  \sum_{i : x_i = 1} w_i \right\rfloor \ge \sum_{i : x_i = 1} \lfloor w_i \rfloor = \lfloor w \rfloor^T x.$$

Therefore $x \in \mathcal{X}_2$, and this shows that $\mathcal{X}_1 \subseteq \mathcal{X}_2$.