If X reduces to a problem in NP, is X in NP?
The answer depends on what you mean by the notation $\le_p$.


If $\le_p$ refers to a Karp reduction (i.e., a many-one reduction), then the answer is "Yes".  It follows from the definition of NP.  See What is the definition of P, NP, NP-complete and NP-hard? to learn more.
Assuming that $\le_p$ refers to a Cook reduction (also known as a polynomial-time Turing reduction), then the answer is "We don't know for sure, but we think not".

In particular, let $X$ be the problem TAUTOLOGY, and $Y$ be the problem SAT.  Then $X \le_p Y$ (we can transform an instance of TAUTOLOGY to an instance of SAT; if you then flip the answer from an algorithm to solve SAT, the result will be the answer to TAUTOLOGY).  We know that SAT is in NP.  However $X$ is co-NP-complete, so $X$ is not likely to be in NP: if $X$ is in NP, then it follows that NP = co-NP, which would be a surprise.  While we can't prove NP $\ne$ co-NP, many/most complexity theorists expect that NP $\ne$ co-NP is true, and from that it would follow that the answer to your question is "No" -- assuming you are referring to Cook reductions.


To learn more about the difference between these two types of reductions and why both notions exist, take a look at the following questions:


What is the difference between these terms?
Can one show NP-hardness by Turing reductions?
https://cstheory.stackexchange.com/q/138/5038
https://cstheory.stackexchange.com/q/686/5038