How do I prove that if A is a $n × n$ matrix with integer entries, then $det(A)$ is an integer? My proof: 
if A is an n x n matrix with integer entries, then this means that every entry of A contains an integer. Therefore, any integer operated on by another integer is also an integer. A determinant is calculated by a series of integer operations on a matrix. Therefore, det(a) is also an integer.  
Is that a sufficient proof? 

Your approach is correct. You could extend it by giving a definition for the determinant and pointing out the closedness of the involved operations.
One such definition defines the determinant of a matrix as alternating multi-linear form of the column vectors of that matrix. This definition involves only additions and multiplications, so you stay within $\mathbb{Z}$.
$$
\DeclareMathOperator{sgn}{sgn}
\DeclareMathOperator{det}{det}
\det(A) = \det(a_1,\dotsc,a_n) 
= \sum_{\pi \in S_n} \sgn(\pi) \, a_{1\pi(1)} a_{2\pi(2)} \dotsb a_{n\pi(n)} 
$$
