Separable Algebra The definition of a separable algebra from the wiki page is

An algebra $A$ over the field $k$ is separable if for every field
  extension $L/k$ the algebra $A\otimes L/k$ is semisimple

Taking $A$ to be a module over a field, i.e. a vector space, and the fact that you can always view the field extension $L/k$ as a vector space (e.g. $L=\mathbb{C}$ and $k=\mathbb{R}$) we get that $A\otimes L/k$ is also a vector space.  Further, we know that every vector space is semi simple.  So my question is, what is an example of an algebra over a field that is not separable according to this definition?
 A: 1. Notions of semisimplicity
In different contexts there are different notions of semisimplicity. I think, that is where your confusion was coming from:

*

*We call a vector space semisimple if it is a direct sum of simple vector spaces.
A simple vector space $V$ is a vector space whose only subspaces are $(0_V)$ and $V$.
Now, this is a fairly useless definition. As every vector space has a basis (in the infinite-dimensional case evoke AoC), every vector space is semisimple (as you observed).

*More generally, we call an $A$-module over a $k$-algebra $A$ semisimple if it is a direct sum of simple $A$-modules. A simple $A$-module $M$ is one that has only $(0_M)$ and $M$ as submodules.
(A $k$-vector space is equivalently a $k$-module over the $k$-algebra $k$. Hence, this definition of semisimplicity is indeed a generalization of the first.)

*As a special case we have the notion of a semisimple $k$-algebra: One that is semisimple as a left module over itself. Before you ask, yes, one could replace "left module" by "right module" here. But that these definitions are equivalent is apparently a non-trivial result.

Let us apply these notions to your question.
Let $A$ be an algebra over a field $k$ and $L/k$ a field extension. Consider the $L$-algebra $A \otimes_k L$. Of course it is semisimple as a vector space (see above). However, it might not be semisimple as an algebra.
2. Constructing a non-separable algebra
Let me give you an example of a non-separable algebra.
Every separable algebra over a field is semisimple (just choose $k=L$, and use the identification $A \otimes_k k \cong A$). Hence, it suffices to look for a non-semisimple algebra. There are quite a few of them.
For a non-standard example consider the Taft Hopf algebra. Let $N\neq 1$ be a positive integer. Let $k$ be a field of characteristic zero such that it has a $N$-th primitive root of unity $\zeta$ (for example, $\mathbb C=k$).
Consider the algebra $H$ over $k$ generated by two elements $g$ and $x$ subject to the relations $x^N=0$, $g^N=1$ and $xg=\zeta gx$. It is $N^2$-dimensional with a basis $g^ix^j$ for $0\leq i,j \leq N-1$.
A well known theorem by Larson and Radford from 1988 states that a finite-dimensional $k$-Hopf algebra over a field $k$ of characteristic zero is semisimple (as an algebra) if and only if $S^2=id$ for its antipode $S$.
It turns out that $H$ can be made into a Hopf algebra with the following antipode: $S_H: H \rightarrow H; g \mapsto g^{-1}, x \mapsto -xg^{-1}.$ However: $${S_H}^2(x)=S_H(-xg^{-1})=-S_H(g^{-1})S_H(x)=gxg^{-1}.$$ Here we have used that the antipode is a linear map and an antialgebra homomorphism.
Because $N\neq 1$ we have just shown that ${S_H}^2\neq id_H$. Hence, the algebra $H$ is not semisimple, and therefore not separable either.
3. On a side note
By the way, various notions of semisimplicity are unified in category theory.
In an abelian category there is the concept of simple objects and semisimple objects. There is an nLab article on these notions.
