Is it a sufficient condition to be in NP? Suppose the following situation.

You have a decision problem $D$.
You know that $SAT$ is $NP$-complete.
You know that $D\leq_p SAT$.

Can you conclude that $D\in NP$?
I think it's true because it means that $D$ is at least as hard as any other problem in $NP$.
As I know I am not able to post questions to answer with yes/no:
If it's true, can you give a better explanation? If it's false, can you argue why is so?
PD: it's not homework.
You can obtain a non-deterministic polynomial-time Turing machine $T$ for $D$ as follows:

Since $D \le_p SAT$, there is a polynomial-time Turing machine $T'$ that maps any instance of $D$ into an instance of $SAT$. Simulate the $T'$ on $x$, where $x$ is the input of $T$.
Use any non-deterministic polynomial-time algorithm to decide SAT (e.g., guess the value of all the variables in the SAT formula and check if the formula is satisfied). By definition of Karp reduction, $T$ will accept if and only if $x \in D$.

This shows that $D \in \mathsf{NP}$.