You ask for an instructive example, so I'll be long winded.

Suppose $(x_n)$ is basis for a normed space $(X,\|\cdot\|)$. The partial sum projections $S_n$ are well defined but might be discontinuous. Define a new, larger, norm on $X$ by $|x| := \sup_n \|S_n x\|$. Then $(x_n)$ is a Schauder basis for $(X,|\cdot|)$. Note that this is the first step in the proof that every basis for a Banach space is a Schauder basis. The harder part of the proof is to show that $(X,|\cdot|)$ is complete when $(X,\|\cdot\|$ is complete. When $(X,\|\cdot\|)$ is not complete, $(X,|\cdot|)$ may not be complete, but if you are building a basis for $(X,\|\cdot\|)$ that is not a Schauder basis, it would be nice to have one such that the basis is a Schauder basis for some natural complete norm on $X$. Also, you would like your non Schauder basis to be in a some natural normed space; let's say an inner product space that is a non closed subspace of $\ell_2$ under the usual $\ell_2$ norm, $\|\cdot\|_2$. Well, $\ell_2$ has lots of non closed subspaces that are complete under a natural norm, the most used being $\ell_1$ under its norm $\|\cdot\|_1$. It should not be too hard to find a Schauder basis for $\ell_1$ such that at least one of the biorthogonal functionals is not in $\ell_2$. Perhaps the simplest such example is $x_1 = e_1$ and $x_n = e_1+e_n$ when $n>1$. (Here $(e_n)$ is the standard unit vector basis.) Notice that the biorthogonal functionals in $\ell_\infty = \ell_1^*$ are given by $x_1^* =(1,-1,-1,\dots)$ and, for $n>1$, $x_n^* = (0,0,\dots,1,0,\dots)$, where the $1$ is in the nth coordinate. Only $x_1^*$ is not continuous in the $\ell_2$ norm--this makes it easy to verify that $(x_n)$ is a basis for $(\ell_1,\|\cdot\|_2$).