Let $e_1 = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}$, and so on, where $e_i$ is a vector of all zeros, except for a $1$ in the $i^{\mathrm{th}}$ place. Since $A$ is positive definite, then $x^T A x > 0$ for any non-zero vector $x \in \Bbb R^n$. Then, $e_1^T A e_1 > 0$, and likewise for $e_2, e_3$ and so on.
If the $i^{\mathrm{th}}$ diagonal entry of $A$ was not positive, $a_{ii} < 0$, then $e_i^T A e_i = 0\cdot a_{11}\cdot 0 + 1\cdot a_{12}\cdot 0 + \cdots + 1\cdot a_{ii}\cdot 1 + \cdots + 0\cdot a_{nn} \cdot 0$, since $e_i$ has zeros everywhere but in the $i^{\rm th}$ spot.
Thus, what would happen if $a_{ii}$ was positive?
I doubt that this is the best approach since it basically only uses the definition of positive semidefinite (PSD) matrices and requires a fact about the pointwise product of PSD matrices, but maybe it will give you some ideas.
We have that,
$$
\sin(x_i-x_j)^2 = \frac{1}{2} - \frac{1}{2}\cos(2(x_i-x_i))
= \frac{1}{2} - \frac{1}{2}\big[ \cos(2x_i)\cos(2x_j) + \sin(2x_i)\sin(2x_j) \big]
$$
Therefore we can write,
$$
A_{i,j} = e^{-\lambda/2}\exp\left(\frac{\lambda}{2}\cos(2x_i)\cos(2x_j) +\frac{\lambda}{2}\sin(2x_i)\sin(2x_j) \right)
$$
Now, note that the matrix,
$$
B_{i,j} = \frac{\lambda}{2}\cos(2x_i)\cos(2x_j) + \frac{\lambda}{2}\sin(2x_i)\sin(2x_j)
$$
is the sum of two rank-1 outer products and therefore PSD.
Note: Maybe for your needs its sufficient to show the part of the kernel in the exponent is the sum of separable functions, in which case you are done.
Using Taylor expansion we have,
$$
A_{i,j}
%= e^{-\lambda/2} \sum_{k=0}^{\infty} \frac{1}{k!} \left( \frac{\lambda}{2}\cos(2x_i)\cos(2x_j) + \frac{\lambda}{2}\sin(2x_i)\sin(2x_j) \right)^k
= e^{-\lambda/2} \sum_{k=0}^{\infty} \frac{1}{k!} B_{i,j}^k
$$
So in matrix form, using $\circ$ to denote the poitwise product of matrices,
$$
A = e^{-\lambda/2} \left( I + B + \frac{1}{2!} B\circ B + \frac{1}{3!} B\circ B\circ B + \cdots \right)
$$
Finally, we claim that this matrix is PSD.
First, note that the pointwise product of two PSD matrices is again PSD. Therefore, each of the terms $B\circ B\circ \cdots \circ B$ is PSD. The sum of PSD matrices is also PSD.
This is an infinite series of pointwise additions and products of PSD matrices. If this converges (which you know it does), it will also be PSD.
Best Answer
The radial basis function $\phi(y) = \exp(-\alpha\sqrt{y})$ is completely monotone on $[0,\infty)$. Hence, by Schoenberg's interpolation theorem (see chapter 15 of A Course in Approximation Theory by Cheney and Light or sec. 2.5 of this book chapter, for instance), $\phi(\|\cdot\|^2)$ is strictly positive definite. Therefore $k(x,z)=\phi(\|x-z\|^2)$ is a kernel and $K$ is positive definite when the data points $x_1,\ldots,x_n$ are distinct (or positive semidefinite otherwise).
Alternatively, $K$ may be viewed as the covariance matrix for two Ornstein-Uhlenbeck processes. Hence it is positive semidefinite.