Q:

Subdividing the sides of a square to approximate a circle

I have read that you can approximate a sphere from an icosahedron by subdividing the faces and normalizing the points created by the subdivisions.  Please see the accepted answer of this question: Algorithm for creating spheres. 
Wouldn't it follow that you could create a circle by subdividing the sides of a square and normalizing the points?  This doesn't seem to work. 
Could someone explain to me why? 

A:

Short answer: Your error is that the norm of a vector $(x,y)$ is not $x+y$, or $\lvert x\rvert+\lvert y\rvert$, but rather $\sqrt{x^2+y^2}$. A vector is normalized if its norm is $1$. If you start with $(\frac12,\frac12)$, its norm is $\sqrt{(\frac12)^2+(\frac12)^2} = \frac1{\sqrt2}$, so you have to divide it by $\frac1{\sqrt2}$ to get the normalized vector $(\frac1{\sqrt2},\frac1{\sqrt2})$.
Long answer: Well, in principle there are different kinds of norms one can use, but only one makes sense here. 
In a geometry context, when someone says "norm" without further qualification, they invariably mean the Euclidean norm, or the $2$-norm, which for an $n$-dimensional vector $(x_1, x_2, \ldots, x_n)$ is $\sqrt{x_1^2+x_2^2+\cdots+x_n^2}$. Why?  Because this is how lengths and distances work.  Get a sheet of graph paper and mark the points $(0,0)$ and $(3,4)$. Measure the distance between them with a ruler: it's not $3+4=7$, it's $\sqrt{3^2+4^2}=5$. Or, connect the points $(0,0)$, $(3,0)$, and $(3,4)$: a wild right triangle appears!  Cast Pythagoras to defeat the beast. 
In general, though, a norm is just — roughly speaking — something that tells you how far away from zero a vector is, and there are lots of possible ways of measuring that.  For example, you could just add up the absolute values of the components, $\lvert x_1\rvert+\lvert x_2\rvert+\cdots+\lvert x_n\rvert$, like you thought; this is known as the $1$-norm.  Or you could take just the largest component, $\max(\lvert x_1\rvert,\lvert x_2\rvert,\ldots,\lvert x_n\rvert)$, which gives the maximum norm, or the $\infty$-norm.  The thing is, though, that because different norms measure distance differently, the "unit spheres" made of points at distance $1$ have different shapes.  In particular, the unit sphere for the $1$-norm is a square oriented at $45^\circ$ degrees, which is exactly what you started with, so normalizing the points on it doesn't change anything.