Both of these rely on the intermediate value theorem.
(i) Define one direction normal to $d$ as up. Draw a line $l_0$ "below" $A$, and one line $l_1$ "above" $A$ (possible because $A$ is bounded) (both parallel to $d$),and call the distance between them $r$. You can define a function $f$ from the interval $[0, 1]$ into $\mathbb R$ taking a number $x$ to the area below the line parallel to $d$ and distance $rx$ above $l_1$. We now have $f(0) = 0$, $f(1) = |A|$, and it is continuous, so somewhere it has to equal $|A|/2$. That's where you put your line.
To solve (ii) we need to know that the line is unique. We therefore need to show that $f$ is monotone, and that at the point where $f(x)= |A|/2$, it is strictly increasing.
The first part is simple, since we only move the line in one direction, points of $A$ can only go from being over the line to being under the line as $x$ increases, so the calculated area cannot decrease.
The second part is a bit more tricky. I'll show the "top part", the bottom part is completely analogus. Given the point $x$ where $f(x) = |A|/2$, for any $\epsilon > 0$ (and small enough that $x+\epsilon < 1$) we must show that $f(x+\epsilon) > f(x)$. Since $A$ is open and connected there will be a point $p$ on the line corresponding to $x$ with $p\in A$ which is the centre of a ball contained in $A$ with radius less than $r\epsilon$. Half this ball will be above the $x$-line, but below the $x+\epsilon$-line, thus $f$ has increased between these two points.
We now know that the line is unique, and can carry on with (ii).
(ii) For each possible direction, find the line that divides $A$ into equal parts by (i). Remark that opposite directions yield the same line, since opposite directions are still parallell. Now pick a direction $d$. Define a direction normal to $d$ as up, and now define a function $g$ from the unit interval to $\mathbb R$ as follows: For a number $x$, turn the direction $d$ (and the orientation "up") a total of $\pi x$ radians counterclockwise and let the value of the function be how much area of $B$ is above the line with that direction, dividing $A$ in half. The key remark here is that $f(0) + f(1) = |B|$, since they are both calculated from the same direction $d$, but with the orientation "up" in opposite direction. You therefore have a function from the unit interval to $\mathbb R$ with $f(0) = |B|/2 + a$ and $f(1) = |B|/2 - a$ for some real number $a$. Thus it has to be $|B|/2$ somewhere in between, and you then have your line.