**Theorem.** For a topological space $X$ the following conditions are equivalent:

1) for any continuous map $f:X\times S^2\to\mathbb R^2$ there exists a point $s\in S^2$ such that $f(x,s)=f(x,-s)$ for any $x\in X$;

2) any continuous map $f:X\to \mathbb R$ is constant.

Proof. (1) $\Rightarrow$ (2) Assume that $X$ admits a non-constant map $\hbar :X\to \mathbb R$. We lose no generality assuming that $\{0,1\}\subset \hbar(X)\subset[0,1]$.
Then there are points $x_0,x_1\in X$ such that $\hbar(x_i)=i$ for $i\in\{0,1\}$.

Let $p:S^2\to\mathbb R^2$, $p:(x,y,z)\mapsto (x,y)$, be the projection of the sphere $S^2=\{(x,y,z)\in\mathbb R^3:x^2+y^2+z^2=1\}$ onto the plane.

Let $\varphi:S^2\to S^2$ be any homeomorphism of the sphere $S^2$ such that $p\circ \varphi(0,0,1)\ne p\circ \varphi(0,0,-1)$.

Using the Tietze-Urysohn Theorem, find a continuous map $\psi:[0,1]\times S^2\to\mathbb R^2$ such that $\psi(0,s)=p(s)$ and $\psi(1,x)=p\circ\varphi(s)$ for all $s\in S$. Then the continuous map $$f:X\times S^2\to\mathbb R^2,\;\;f:(x,s)\mapsto \psi(\hbar(x),s),$$ has the following property:

if $f(x_0,s)=f(x_0,-s)$ for some $s\in S^2$, then $s\in\{(0,0,1),(0,0,-1)\}$ and $f(x_1,s)\ne f(x_1,-s)$.

(2) $\Rightarrow$ (1) Assume that each continuous map $X\to\mathbb R$ is constant and take any continuous function $f:X\times S^2\to\mathbb R^2$. Fix any point $x_0\in X$ and using the Borsuk-Ulam Theorem, find a point $s\in S^2$ such that $f(x_0,s)=f(x_0,-s)$.
By our assumption, for every $s\in S^2$ the function $f{\restriction}X\times\{s\}$ is constant. So, for every $x\in X$ we have
$$f(x,s)=f(x_0,s)=f(x_0,-s)=f(x,-s).$$

**Remark.** For examples of regular topological spaces on which all continuous real-valued functions are constant, see page 119 of Engelking's "General Topology".