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Pete's is certainly the right way to look at this problem, but in this example one can argue naively using explicit calculations. One loses no generality by assuming $c=0$ (by replacing $x$ by $x+c$). Then using the duplication formula, one finds that the solutions of $[2]P = (0,0)$ are $P=(uv,uv(u+v))$ where $u$ and $v$ run through the square roots of $-a$ and $-b$ respectively. If $-a$ and $-b$ are squares in $k$ then each $P$ has coordinates in $k$. If one of the $P$ has coordinates in $k$ then they all do: so both $(uv,uv(u+v))$ and $(-uv,-uv(u-v))$ lie in $E(k)$. Thus $uv$, $u+v$ and $u-v$ lie in $k$. Hence $u\in k$ and $v\in k$ so that $-a$ and $-b$ are squares in $k$.