Let the roots of $p(x) = x^2 + dx + a$ be $x_1$ and $x_2$, and the roots of $q(x) = x^2 + cx + b$ be $y_1$ and $y_2$.

Since $p(x)$ has two different real roots, its discriminant is positive:
$$D_p = d^2 - 4a > 0$$
Similarly, $q(x)$ has two different real roots, so
$$D_q = c^2 - 4b > 0$$

Suppose, for contradiction, that some root of $p(x)$ coincides with some root of $q(x)$. That is, there exists $x_0$ such that $p(x_0) = 0$ and $q(x_0) = 0$.

So:
$$x_0^2 + d x_0 + a = 0$$
$$x_0^2 + c x_0 + b = 0$$
Subtracting the second from the first:
$$(d-c)x_0 + (a-b) = 0$$
Since $d > c$, $d-c > 0$.
So:
$$x_0 = \frac{b-a}{d-c}$$

Now, plug $x_0$ into $p(x)$:
$$\left(\frac{b-a}{d-c}\right)^2 + d \left(\frac{b-a}{d-c}\right) + a = 0$$
Multiply both sides by $(d-c)^2$:
$$(b-a)^2 + d(b-a)(d-c) + a(d-c)^2 = 0$$
Expand $d(b-a)(d-c)$:
$$d(b-a)(d-c) = d(b-a)d - d(b-a)c = d^2(b-a) - d c(b-a)$$
So:
$$(b-a)^2 + d^2(b-a) - d c(b-a) + a(d^2 - 2 d c + c^2) = 0$$
Expand $a(d^2 - 2 d c + c^2)$:
$$a d^2 - 2 a d c + a c^2$$
So the equation is:
$$(b-a)^2 + d^2(b-a) - d c(b-a) + a d^2 - 2 a d c + a c^2 = 0$$
Group terms:
$$(b-a)^2 + d^2(b-a) - d c(b-a) + a d^2 - 2 a d c + a c^2 = 0$$

Now, let's try to show that this equation cannot hold for $0 < a < b < c < d$.

Alternatively, consider the following:

The sum of the roots of $p(x)$ is $-d$, and the product is $a$.
The sum of the roots of $q(x)$ is $-c$, and the product is $b$.

Suppose $x$ is a root of both $p(x)$ and $q(x)$.
Then $x^2 + d x + a = 0$ and $x^2 + c x + b = 0$.
Subtract:
$$(d-c)x + (a-b) = 0$$
So $x = \frac{b-a}{d-c}$.

Now, since $p(x)$ has two real roots, $d^2 > 4a$.
Similarly, $c^2 > 4b$.

But $a < b$ and $c < d$.

Now, let's check if $x = \frac{b-a}{d-c}$ can be a root of $p(x)$.
Plug into $p(x)$:
$$\left(\frac{b-a}{d-c}\right)^2 + d \left(\frac{b-a}{d-c}\right) + a = 0$$
Multiply both sides by $(d-c)^2$:
$$(b-a)^2 + d(b-a)(d-c) + a(d-c)^2 = 0$$
But $b-a > 0$, $d-c > 0$, $a > 0$.
All terms are positive, so the sum cannot be zero.

Therefore, $p(x)$ and $q(x)$ cannot have a common root.

Now, suppose $p(x)$ has a double root, i.e., $x_1 = x_2$.
But the discriminant is positive, so roots are distinct.
Similarly for $q(x)$.

Suppose $p(x)$ has two roots, $x_1$ and $x_2$, and $q(x)$ has two roots, $y_1$ and $y_2$.
Suppose $x_1 = y_1$.
Then, as above, $x_1 = \frac{b-a}{d-c}$, but this is not possible as shown above.

Suppose $x_1 = y_2$ or $x_2 = y_1$ or $x_2 = y_2$.
The same argument applies.

Therefore, all four roots are mutually different.