Consider the plane ${\mathbb R}^2$ with points $z:=(x,y)$. The euclidean metric is given by d_{\rm eucl}(z,z'):=\sqrt{(x-x')^2+(y-y')^2}\ ; it has the usual euclidean circle of radius $1$ as its unit sphere. On the other hand, your pixel counting corresponds to the so-called $l^\infty$-metric d_{\rm pix}(z,z'):=\max\{|x-x'|,|y-y'|\}\ ; the unit sphere in this case is the square with vertices $(\pm 1,\pm1)$.
To measure the lengths of a curve $\gamma$, e.g., the diagonal of a large square, one divides the curve $\gamma$ into $N\gg1$ little "segments" and covers each segment with an appropriately scaled unit disk. (This is the idea of the so-called Hausdorff measure.) In this way, using $d_{\rm eucl}$ one finds that the diagonal of the unit square has length $\sqrt{2}$, and using $d_{\rm pix}$ instead one arrives at length $1$.
To put it differently: If your pixels were little circles that can be moved around at will, the limiting process you describe would give $\sqrt{2}$ as length of said diagonal.
Added later:
Both metrics induce the same topology (the "usual" topology) in the plane. A sequence $(z_n)_{n\geq0}$ converges to some point $z^*\in{\mathbb R}^2$ (or not) whether you measure distances by means of $d_{\rm eucl}$ or $d_{\rm pix}$. The essential difference between the two metrics is the following: Both metrics are translation invariant and behave in the expected way under scaling. But $d_{\rm eucl}$ is also invariant under rotation, which $d_{\rm pix}$ is not. As a consequence $d_{\rm eucl}$ allows of a much richer geometry (which you learn in high school) than $d_{\rm pix}$.