An inner product on a real or complex vector space $V$ is a bilinear map $V\times V\to K$ (where $K$ is the ground field, either $\bf R$ or $\bf C$), that satisfies conjugate symmetry and positive-definiteness; for the second property to make sense we have to realize that $\langle x,x\rangle$ is real for all $x\in V$ (this follows from the first property actually), so it makes sense to say it is nonnegative.
The map $\|x\|=\langle x,x\rangle^{1/2}$ will in fact be a vector space norm, and this norm induces a metric via the formula $d(u,v)=\|u-v\|$. The nontrivial part of checking these facts is using Cauchy-Schwarz for establishing the triangle inequality (fix an orthogonal basis to do it in).
It is not possible to define an inner product on a vector space over a field of positive characteristic, by definition. It is, however, possible to define bilinear forms $(\cdot,\cdot):{\bf F}_q^n\times{\bf F}_q^n\to{\bf F}_q$, and two vectors are orthogonal with respect to it if $(a,b)=0$. In most cases of positive characteristic and dimension greater than one it is possible to find an $x$ which is orthogonal to itself under the usual coordinate-determined dot product (this is actually an interesting number-theoretic question: over which finite fields and numbers $n$ do there exist $n$ scalars not all zero whose squares sum to zero?)
It is also not possible to define a metric $X\times X\to {\bf F}$ where $\bf F$ is a field of positive characteristic: by definition in the first place it would have to take real values, but furthermore it could not satisfy the triangle inequality since there can be no ordering in positive characteristic.
Not every vector space is an inner product space because not every norm satisfies the parallelogram law. As classic counterexamples, consider the the spaces $\mathbb{R}^n$ under the $1$-norm (aka taxicab norm) and the $\infty$-norm (aka maximum/supremum norm).
If you are given an inner-product space (aka Hilbert space), then there is indeed a strong connection between the dual space and the inner product. This result is known as the Riesz representation theorem. Note, however, that his does not mean that we've "defined the same thing twice". The dual space is the set of all linear mappings to the scalar field, whereas the inner product of an inner product space is a particular (bilinear) map on two vectors to the scalar field.
Best Answer
You have the right type of ideas, but missing a couple of details that I consider important. Here's how I see it:
To be a metric space, you need no structure on the set, and all you get is a distance with the triangle inequality (in terms of how it has to be constructed) and positive definiteness.
To be either a normed space, or an inner product space, you need a vector space over $\mathbb{R}$ (or $\mathbb{C}$ with some slight adjustments).
A norm gives you a weight which has the triangle inequality in terms of the addition on the space, and respects the multiplication by scalars. This can be used to induced a metric (but not all metrics arise in this way, because there is metrics that don't respect the sums or multiplication).
An inner product gives you the above, except this time the weight is induced by a positive-definite symmetric bilinear form. The fact that it corresponds to this symmetric bilinear form is what gives you the parallelogram identity and other things that exist in inner product spaces but not all normed spaces. (but not all norms arise in this way, because there is norms not induced by a positive-definite bilinear form)