What is the difference between a Hilbert space and Euclidean space? According to Wikipedia, 

Hilbert space [...] extends the methods of vector algebra and calculus from the two-dimensional Euclidean plane and three-dimensional space to spaces with any finite or infinite number of dimensions

However, the article on Euclidean space states already refers to 

the n-dimensional Euclidean space.

This would imply that Hilbert space and Euclidean space are synonymous, which seems silly.
What exactly is the difference between Hilbert space and Euclidean space? What would be an example of a non-Euclidean Hilbert space?
 A: A Hilbert space essentially is also a generalization of Euclidean spaces with infinite dimension.

Note: this answer is just to give an intuitive idea of this generalization, and to consider infinite-dimensional spaces with a scalar product that they are complete with respect to metric induced by the norm.
Clearly, there are finite-dimensional Hilbert spaces, as $\mathbb{R}^n$, with the standard scalar product and Euclidean metric.
A: Hilbert space: a vector space together with an inner product, which is a Banach space with respect to the norm induced by the inner product 
Euclidean space: a subset of $\mathbb R^n$ for some whole number $n$
A non-euclidean Hilbert space: $\ell_2(\mathbb R)$, the space of square summable real sequences, with the inner product $((x_n),(y_n)) = \sum_{n=1}^{\infty}x_n y_n$
A: An Euclidean space is a normed linear space, that is, it has a norm and its elements are linear functions.
An Euclidean space has an inner product (scalar product):
$$(x_\alpha,x_\beta)=0$$ for orthogonal elements
and
$$(x_\alpha,x_\alpha)=1$$
This scalar product must satisfy the following:
$$
\begin{array}
f\bullet\  being\ not\ negative (x,x>0)\\
\bullet\ being\ symmetric (x,y)=(y,x)\\
\bullet\ being\ linear: \lambda x, y)=\lambda(x,y)\\
\bullet\ (x,y+z)=(x,y)+(x,z)
\end{array}
$$
We say therefore that any space equipped with an inner product satisfying these properties is called Euclidean space.
Euclidean space is therefore a general term for a normed linear space with an inner product, where the norm can be
$$d(x,y)=||x-y||=\sqrt{(x-y)^2}$$
The norm is a distance in the linear normed space, between two points.
Now, a Hilbert space is an example of an Euclidean space. It has an inner product, however it is of infinite many dimensions. For instance, considering a particle in quantum mechanics as a dimension, one can form a Hilbert space of 10 dimensions for a system of 10 particles, where the particle must satisfy the orthogonality principle of the inner product stated at the top. Therefore, all the wavefunctions for the particles given above give an inner product equal to zero, since they have the same angle between one another. Have  a look at the atomic orbital chart and see how they have the same angle preserved between each wavefunction.
This wavefunction model is then said to satisfy the conditions for the inner product, and you can have infinitely many of them. If you used an Euclidean space to describe a system of particles, you were restricted to 3 particles, that is, the three dimensions Euclidean space is restricted to. Therefore, you will see that Hilbert space is commonly used in quantum mechanics, with all its properties of the Euclidean space, but with its infinitely many dimensions.
A: *

*A Hilbert space does not have to be infinite dimensional (it could be).

*The Euclidean space is an example of a finite dimensional (n- dimensional) Hilbert space where the scalar field is the set of real numbers, i.e., $\mathbb{R}^n$.

*It is best to leave out quantum mechanical discussions out of this since it will generally confuse the issue. But, I feel compelled, for the sake of students in QM, to correct Luther4.
(i) In quantum mechanics the wavefunction (state vector) of a single particle system is an element of a Hilbert space. The scalar field used in QM is the set of complex numbers. In general the Hilbert space of a single particle system is infinite dimensional. In many example, a finite dimensional subset is sufficient to describe the system.
(ii) There is no connection between the number of particles and the dimension of the Hilbert space used in QM. A m-particle system can be described by a cartesian product of m single particle Hilbert spaces. If quantum indistinguishability of the particles is imposed (Fermions Bosons etc,), a subset of the Hilbert space that satisfies certain symmetry with regards to particle exchange is used.

