What you want to prove is not necessarily true. Here is a random counterexample:
$$
D=\left[\begin{array}{c|c}A&C\\ \hline C^T&B\end{array}\right]
=\left[\begin{array}{rr|rr}
2&-3&1&0\\
2&2&0&-1\\
\hline
1&0&2&1\\
0&-1&-2&2
\end{array}\right].
$$
According to your definition, a matrix is "positive definite" if and only if its symmetric part is positive definite in the conventional sense. Now, WolframAlpha reckons that the eigenvalues of
$$
D+D^T=\left[\begin{array}{rr|rr}
4&-1&2&0\\
-1&4&0&-2\\
\hline
2&0&4&-1\\
0&-2&-1&4
\end{array}\right]
$$
are $4-\sqrt{5}$ and $4+\sqrt{5}$ (each of multiplicity 2). Hence $D$ is "positive definite". WolframAlpha also reckons that $\det(D)=61$. However,
$$\det(A)\det(B)=10\times6=60<\det(D).$$
To test for positive definiteness, as long as the principal submatrices are nested (and we have submatrices of all sizes, of course), they don't need to be the leading ones or the trailing ones. This is because by relabelling the rows and columns, any nested sequence of principal submatrices can be turned into a sequence of leading principal submatrices.
E.g. for any $S\subseteq\{1,2,3,4,5\}$, let $A(S)$ denotes the principal submatrix $(a_{ij})_{i,j\in S}$. Then $A(\{4\}),\,A(\{2,4\}),\,A(\{1,2,4\}),\,A(\{1,2,4,5\})$ and $A$ form a nested sequence of principal submatrices, and $A$ is positive definite if and only if the determinants of these submatrices are all positive.
In your example, $A_1=A(\{1\},\,B_2=A(\{4,5\}),\,A_3=A(\{1,2,3\}),\,A_4=A(\{1,2,3,4\})$ and $B_5=A$ do not form a nested sequence of principal submatrices. Even if their determinants are all positive, $A$ is not necessarily positive definite. For a counterexample, consider $A=\operatorname{diag}(1,-1,-1,1,1)$.
By the way:
- You lack vocabularies. I think you need to look up the meanings of the words "leading/trailing" and principal in your textbook.
- The test has a name. It is called Sylvester's criterion.
- You seem to have misunderstood how to use Sylvester's criterion for semidefiniteness. To verify that a Hermitian matrix is positive semidefinite, you need to show that all principal minors are nonnegative. It does not suffice to show that all leading principal minors are nonnegative. For a counterexample, consider $A=\operatorname{diag}(0,1,-1,0)$. All leading principal minors of this $A$ are zero (hence nonnegative), but $A$ is indefinite.
Best Answer
The easiest way (in my opinion) to see that the matrix is positive definite is to check two things: is the matrix symmetric (yes and easily checkable) and are all the eigenvalues positive (yes, but less obvious).
You can quickly see that the eigenvalues are all positive using the Gershgorin circle theorem (https://en.wikipedia.org/wiki/Gershgorin_circle_theorem). In short, all the eigenvalues (for your matrix) live in (at least) one of the balls: $B(25, 20)$ (ball centered at 25 of radius 20), $B(18, 15)$, and $B(11, 5)$. All of those balls are strictly positive, so all your eigenvalues are positive.
Symmetry combined with all positive eigenvalues implies positive definite.