Let K = Zp(x,y) for any prime p, say 5.
All fifth powers have exponents divisible by 5, so x has no fifth root in K. Let u be the fifth root of x, and let v be the fifth root of y. Let F = K(u,v). Since K is generated by purely inseparable elements it is a purely inseparable extension over K.
Consider the fifth powers of the elements in K(u). Remember, we can take the fifth power by multiplying each exponent by 5. This is the frobenius homomorphism. So u → x, u2 → x2, and so on. At the same time, y → y5, and all exponents on y are divisible by 5. This means K(u) does not include v, since v5 → y. The roots u and v are independent.
Does v still have dimension 5 over K(u)? It's dimension is at most 5, since it is the fifth root of y. And v is purely inseparable over K(u), so it has to be the fifth root of something in K(u), hence its dimension is still 5, and F/K has dimension 25.
Suppose K(w) = K(u,v), so that w25 lies in K. Yet everything in K(u,v), when raise to the fifth power, lies in K, so w5 lies in K, and w cannot come from an irreducible polynomial of degree 25. The extension F/K is not simple. This means it is not separable, but we already knew that. It also means there are infinitely many intermediate extensions, as we shall see below.
Every intermediate field is purely inseparable with dimension 5. The fields K(u) and K(v) are two examples.
For any integer j, consider the extension K(u+vxj). that is, adjoin the fifth root of x+yx5j. All such extensions are purely inseparable with dimension 5. Suppose K(u+vxi) also contains u+vxj. It then contains the difference, (xi-xj)v, and hence v, and hence u. The extension is F, and cannot have dimension 5, which is a contradiction. All such extensions are distinct, and F/K has an infinite number of intermediate fields.