Question: I have two statements 

> 
> "A set X is nowhere dense"
> 
> 
> "A set X has no interior"
> 
> 
> 

Are these both equivalent statement?

Answer: 
No, $\mathbb{Q}\subset\mathbb{R}$ has an empty interior, but is dense in $\mathbb{R}$.
For possible future use by others, it might be helpful to dissect these notions a bit more systematically in a metric space setting to see what makes them different. In particular, this will show how being nowhere dense is a super-strong way of having empty interior.

$B(x,\epsilon)$ denotes the open ball centered at $x$ with radius $\epsilon,$ where $\epsilon > 0$ is assumed.

$X$ has empty interior
======================

*This is equivalent to each of the following:*

$X$ has no interior points.

For each $x$ in the space, $x$ is not an interior point of $X.$

For each $x$ in the space, there is no nonempty open set containing $x$ that is a subset of $X.$

For each $x$ in the space, there is no ball $B(x,\epsilon)$ that is a subset of $X.$

For each $x$ in the space, each ball $B(x,\epsilon)$ contains **at least one point that belongs to the complement of** $X.$

$X$ is nowhere dense
====================

*This is equivalent to each of the following:*

$X$ is not dense in each nonempty open set.

$X$ is not dense in each ball $B(x,\epsilon).$

For each $x$ in the space, each ball $B(x,\epsilon)$ is NOT densely filled with points from $X.$

For each $x$ in the space, each ball $B(x,\epsilon)$ contains at least one nonempty open set that is a subset of the complement of $X.$

For each $x$ in the space, each ball $B(x,\epsilon)$ contains **at least one ball all of whose points belong to the complement of** $X.$