Here's a bit of a rough sketch I hope works, at least in $\mathbb{R}^3$:
Each $s\in S^2$ can be identified with a unit vector in $\mathbb{R}^3$. For each $s\in S^2$, you can define $P_i^s$ to be the unique (edit: the plane actually need not be unique, but can by chosen in a systematic way as pointed out in the comments below) plane with normal vector $s$ which divides $A_i$ into two pieces of equal measure. At least in $\mathbb{R}^3$, I think it's intuitively clear such $P_i^s$ exist. Since the $A_i$ are compact, they are bounded, so by continuously sliding an affine plane along the line determined by $s$, at some point none of $A_i$ on one side of the plane, but as we slide along, the amount of measure of $A_i$ on that side of the plane increases from none to all. So by the intermediate value theorem, at some point the plane must divide the measure of $A_i$ equally.
So let $d_1\colon S^2\to\mathbb{R}$ be the continuous function for which $d_1(s)$ which measures the distance from $P_3^s$ to $P_1^s$, where the distance is positive (nonnegative, I guess) if we travel in the $s$ direction to get from $P_3^s$ to $P_1^s$, and negative (nonpositive, I guess) if we travel in the $-s$ direction to get from $P_3^s$ to $P_1^s$. Likewise define $d_2\colon S^2\to\mathbb{R}$ for the distance from $P_3^s$ to $P_2^s$.
Then define
$$
\varphi\colon S^2\to\mathbb{R}^2\colon s\mapsto (d_1(s),d_2(s)).
$$
This is a continuous map, and $\varphi(-s)=-\varphi(s)$ since changing the direction of $s$ changes the sign of the distance between the planes found above. By Borsuk-Ulam, there exists $s_0\in S^2$ such that $\varphi(s_0)=\varphi(-s_0)$, which means $d_1(s_0)=-d_1(s_0)$, so $d_1(s_0)=0$, and likewise $d_2(s_0)=0$. This means that the distances from $P_1^{s_0}$ and $P_2^{s_0}$ to $P_3^{s_0}$ are both $0$, which means all three planes are the same. So there is a single plane which bisects $A_1$, $A_2$, and $A_3$.
$(2)\Rightarrow(3)$ This is just topology. Suppose first that $S^n=A_1\cup A_2$ with each $A_i$ open. We have $(S^n\setminus A_1)\cap (S^n\setminus A_2)=S^n\setminus(A_1\cup A_2)=\emptyset$, so since each $S^n\setminus A_i$ is closed we can use normality to find an open set $U$ with $S^n\setminus A_1\subseteq U\subseteq \overline U\subseteq A_2$. Putting $B_1=S^n\setminus U$ and $B_2=\overline U$ we have $S^n=B_1\cup B_2$ with each $B_i$ a closed set satisfying $B_i\subseteq A_i$.
Now suppose $S^n=A_1\cup\dots\cup A_{n+1}$ with each $A_i$ open. Inductively repeating the above procedure we find closed $B_1,\dots,B_{n+1}\subseteq S^n$ with $B_i\subseteq A_i$ and $S^n=B_1\cup\dots\cup B_{n+1}$. By hypothesis some $B_i$ contains an antipodal point, and hence so does some $A_i$. $\square$
$(3)\Rightarrow(1)$ Let $f:S^n\rightarrow \mathbb{R}^n$ be given and put $g(x)=f(x)-f(-x)$ so that $g(x)=0$ if and only if $f(x)=-f(x)$. Note that $g(-x)=-g(x)$ for all $x\in S^n$. To establish a contradiction we suppose that $0\not\in g(S^n)$.
Take the $n$ open sets $U_i=\{x_i>0\}\subseteq\mathbb{R}^n$, $i=1,\dots,n$, and add to this collection the set $U_{n+1}=\{x_i<0\;\text{for all}\;i=1,\dots,n\}$. Because of the supposition we have $g(S^n)\subseteq\bigcup^{n=1}_{i=1}U_i$.
Letting $A_i=g^{-1}(U_i)$ we can now write $S^n=A_1\cup\dots\cup A_{n+1}$ as a union of $n+1$ open sets. By assumption one of the sets contains an antipodal pair. But this means that there is $x\in S^n$ such that $g(x),g(-x)=-g(x)\in U_i$ for some $i$, and this is absurd. Thus we have a contradiction to the assumption that $g$ is nonvanishing. $\square$.
Best Answer
It is true, and relatively easy to prove given the above result.
If $S^2 = A_1 \cup A_2$, where $A_1, A_2$ are closed, then choose some closed $A_3$ that definitely doesn't contain a pair of antipodal points (e.g. let $A_3$ be a singleton set). Then $S^2 = A_1 \cup A_2 \cup A_3$, and $A_1, A_2, A_3$ are closed, so by the above result, $A_1$, $A_2$, or $A_3$ must contain a pair of antipodal points. Since $A_3$ definitely does not contain such a pair, it must be true of $A_1$ or $A_2$.