I got a rough proof. Suppose otherwise, then in each quadrant, the area of $K$ in the quadrant is more than 1. Now consider a point $(x_1,y_1)$ in $K\cap Q_1$ with $x_1\neq 0$ such that $\frac{y_1-1}{x_1}$ is maximum. Then let $l_1$ be the line connecting $(x_1,y_1)$ and $(0,1)$. Then by definition, the region $K\cap Q_1$ is below $l_1$. Now suppose there is a point in $K\cap Q_2$ above $l_1$, then since $(0,0)\in K$, $(0,1)$ is also in $K$, contradiction, thus $K\cap Q_2$ is below $l_1$. Now consider a point $(x_2,y_2)\in K\cap Q_2$ such that $y_2\neq 0$ and $\frac{-x_2-1}{y_2}$ is maximum. Again define $l_2$ to be the line connecting $(x_2,y_2)$ and $(-1,0)$ and $K\cap Q_2,K\cap Q_3$ is to the left of $l_2$. Again let $(x_3,y_3)\in K\cap Q_3$ with $x_3\neq 0,\frac{y_3+1}{x_3}$ maxed and let $(x_4,y_4)\in K\cap Q_4$ with $y_3\neq 0,\frac{x_4-1}{-y_3}$ maxed, then similarly $K\cap Q_3,K\cap Q_4$ lies above $l_3$ and $K\cap Q_4,K\cap Q_1$ lies to the left of $l_4$. Thus $K$ is bounded by the lines $l_1,l_2,l_3,l_4$, which forms a quadrilateral with its 4 vertex in each quadrant. WLOG, let the angle at the vertex in quadrant 1 (let it be $(a,b)$) be $\alpha\geq 90$, but the area of $K\cap Q_1$ is the maximum when $a=b$, since the area of the triangle $(0,1),(1,0),(a,b)$ is max, but it is easy to find that its area is $\leq 1$, a contradiction. So we are done.