Let A be a $n\times n$ non-singular symmetric matrix with entries in $(0,\infty)$. Let $A$ be a $n\times n$ non-singular symmetric matrix with entries in $(0,\infty)$. Then we can conclude that 
(a)  $|A| > 0 (|A|$ denotes the determinant of A).
(b) $A$ is a positive definite matrix.
(c)  $B = A^2$ is a positive definite matrix.
(d) $C = A^{-1}$ is a matrix with entries in $(0,\infty)$.
I took an example matrix $A=
        \begin{bmatrix}
        1 & 2\\
        2 & 1 \\
        \end{bmatrix}
$, and saw that options (a), (b) and (d) are wrong. 
But what is the proper approach for this problem ?

For a, b and d your approach is right you have a counterexample. Let's take a look at c. $A$ symmetric non singular and therefore is diagonalisable with real nonzero eigenvalues. $A^2$ will then have positive eigenvalues (the squares of those of $A$) on top of being symmetric so it is positive definite. Non singular is an essential assumption as $\begin{bmatrix} 1&1\\1&1\end{bmatrix}$ shows.
