= Solution
For one <electron> per square cell, $n_e=1/a^2$, so
$$
\boxed{\frac{k_F}{\pi/a}=\sqrt{\frac2\pi}\simeq0.7979<1.}
$$
Numerically, $\pi/a\simeq2.17\times10^{10}\ {\rm m^{-1}}$ exceeds $k_F\simeq1.73\times10^{10}\ {\rm m^{-1}}$, so the <fermi circle> lies entirely inside the <first Brillouin zone>.
The state-counting explanation is that a spin-degenerate band has two <electron> states per primitive cell. One <electron> per cell therefore occupies half of that band. In the circular free-electron approximation the occupied reciprocal-space area is
$$
\frac{\pi k_F^2}{(2\pi/a)^2}=\frac12.
$$
A disk of that area fits inside this square: its radius is less than the distance to an edge. Half filling alone is not a shape-independent theorem that every possible <Fermi surface> must avoid every zone boundary; the square geometry and circular free-electron dispersion make the statement true here.
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