My query is that, in the above result we need not define the explicit form of the inner product of the states and it seems like any definition for inner product would work.
You need a definition of inner product to know whether $H$ is Hermitian/self-adjoint or not.
Take the vector space $\mathbb C^2$ and consider the two inner products
$$\left<\pmatrix{a\\b},\pmatrix{c\\d}\right>_1 := \overline ac + \overline b d$$
$$\left<\pmatrix{a\\b},\pmatrix{c\\d}\right>_2 := \overline ad + \overline b c$$
Note: the second is not actually an inner product because it is not positive definite. As a result, it does not produce a Hilbert space. However, as fixing this would add substantial algebraic clutter without meaningfully changing the answer, I have elected to keep this "mistake" intact for now.
Let $M:= \pmatrix{w&x\\y&z}$ be a linear operator. If we use inner product $1$ to compute its adjoint, we obtain
$$\left<\pmatrix{a\\b},M\pmatrix{c\\d}\right>_1 = \overline a\big(wc+xd\big)+\overline b\big(yc+zd\big)$$
$$= \overline{\big(\overline wa+\overline yb\big)}c + \overline{\big(\overline xa + \overline zb\big)}d = \left<M^\dagger_1\pmatrix{a\\b},\pmatrix{c\\d}\right>_1$$
$$\implies M^\dagger_1 = \pmatrix{\overline w & \overline y\\\overline x & \overline z}$$
Doing the same thing for the second inner product yields
$$\left<\pmatrix{a\\b},M\pmatrix{c\\d}\right>_2 = \overline a\big(yc+zd\big)+\overline b\big(wc+xd\big)$$
$$= \overline{\big(\overline za + \overline xb\big)}d+\overline{\big(\overline ya+\overline wb\big)}c = \left<M^\dagger_2\pmatrix{a\\b},\pmatrix{c\\d}\right>_2$$
$$\implies M^\dagger_2 = \pmatrix{\overline z & \overline x\\ \overline y & \overline w}$$
Clearly whether $M$ is self-adjoint depends on which inner product we choose. In the former (conventional) case, $M$ is self-adjoint if its diagonal entries are real and its off-diagonal entries are complex conjugates of one another. In the latter case, the condition for self-adjointness is that the diagonal entries are complex conjugates of one another and the off-diagonal entries are real.
The takeaway here is that when we say an operator is self-adjoint, we mean self-adjoint with respect to a particular inner product. All of the subsequent properties of the operator - and in particular, the orthogonality of its eigenspaces - are with respect to the that inner product.
Lastly, at least for finite-dimensional Hilbert spaces, physicists essentially always use the standard inner product $\langle\cdot,\cdot\rangle_1$. However, we are not missing anything by doing so. Note that given any choice of inner product we may choose an orthonormal basis for the Hilbert space; a linear map between these bases constitutes an isometric isomorphism.
In that sense, the two Hilbert spaces which we obtain by equipping $\mathbb C^2$ with the two different inner products are equivalent.
