We define $\lfloor x\rfloor$ by
$$\lfloor x\rfloor \in \mathbb{Z} \land \lfloor x\rfloor \leq x \land( \forall z \in \mathbb{Z}, z\leq x \Rightarrow z\leq\lfloor x\rfloor)$$
Prove or Disprove the following statement:
$$\exists x \in \mathbb{R}, \forall \epsilon>0,\exists\delta>0,\forall w \in \mathbb{R},|x-w|<\delta \Rightarrow|\lfloor x\rfloor - \lfloor w\rfloor| < \epsilon $$
How should I know that if this statement is true or false before I start to prove? I draw a graph, it seems false to me. Because no matter what $x$ I pick, I will always find a $\delta$ so that $|x-w|$ will be in $\delta$ range, but this doesn't guarantee that $|\lfloor x\rfloor - \lfloor w\rfloor| < \epsilon $.
But I have trouble of finding such $x,\delta$ in general
As MPW pointed out (and as is usual, and as more comments agree on that), $\lfloor x \rfloor$ is the usual floor function, largest integer not exceeding $x$. (That is exactly what that formula on top says: $\lfloor x \rfloor$ is an integer, it does not exceed $x$ and there is no larger $z$ with these properties).
So, given that, take $x=1/2$ and $\delta=1/4$ (no matter what is $\epsilon>0$. Then $w$ must be in the interval $(1/4,3/4)$ so $\lfloor w \rfloor = \lfloor x \rfloor = 0$, so $| \lfloor w \rfloor - \lfloor x \rfloor | = 0 < \epsilon$.
Which, just in case it is not clear, is a proof of your statament $$\exists x \in \mathbb{R}, \forall \epsilon>0,\exists\delta>0,\forall w \in \mathbb{R},|x-w|<\delta \Rightarrow|\lfloor x\rfloor - \lfloor w\rfloor| < \epsilon $$