The Borsuk-Ulam theorem says that, given a continuous map $f: S^n \to \Bbb R^n$, there is some point $x \in S^n$ with $f(x)=f(-x)$. There may, of course, be many such points (maybe $f$ is constant!) But I'd still like to show that some "equivariant" point is stable in the following sense.
We start with a map $\varphi: X \times S^n \to \Bbb R^n$, $X$ some compact manifold. We think of this as a continuously varying family of maps $\varphi_x: S^n \to \Bbb R^n$, given by $\varphi_x(p) = \varphi(x,p)$. The Borsuk-Ulam theorem guarantees there is some point $p_x$ with $\varphi_x(p_x) = \varphi_x(-p_x)$.
What I'd like to know is: is there always a continuous function $\psi: X \to S^n$, such that $\varphi_x(\psi(x)) = \varphi_x(-\psi(x))$? (This should be called stability of the equivariant point because given a map $\varphi_x$, under a mild perturbation of $\varphi_x$, we may still find two equivariant points that are close to one another.) Said another way, given a continuous family of maps $S^n \to \Bbb R^n$, can I continuously choose an equivariant point?
It would be good to prove this at first for $X = I$. If one can do that, and prove a relative version of the theorem for $D^n$ relative its boundary (including $n=1$), one can prove it for all finite CW-complexes. I tend to think that it will be true provided it's true for $X = I$. But I can't seem to make any progress on this case.
As stated, the answer is no. Consider the case of $n=1$ so that maps $f: S^n \rightarrow \mathbb{R}^n$ are simply maps $f: [0, 2\pi] \rightarrow \mathbb{R}$ for which $f(0) = f(2\pi)$, and the Borsuk-Ulam theorem states for every such continuous map $f$ there exists $\theta \in [0, 2\pi)$ such that $f(\theta) = f(\theta+\pi)$.
Let $X = [0,1]$ and consider $\varphi: [0,1] \times [0, 2\pi] \rightarrow \mathbb{R}$ given by $$\varphi(t,\theta) = t(1-t) \phi(\theta),$$ where $\phi: [0, 2\pi] \rightarrow \mathbb{R}$ is supported in $[0, \pi]$ and defined on its support by $$\phi(\theta) = |\theta-\pi/2|-\pi/2 \text{ (for }0 \le \theta \le \pi).$$ The graph of $\phi$ is given by
Notice that $\varphi : [0,1] \times [0, 2\pi] \rightarrow \mathbb{R}$ is continuous and in fact defines a homotopy from the constant $0$ function to itself. Moreover, $$\{ (t, \theta) \in [0,1] \times [0,2\pi] : \varphi(t,\theta) = \varphi(t,\theta + \pi) \} =\big(\{0,1\}\times[0, 2\pi]\big) \cup\big([0,1] \times \{0, \pi, 2\pi\}\big).$$ Now, of course, we can continuously choose an equivariant point for $\varphi$. Namely, take $\psi :[0,1] \rightarrow [0,2\pi]$ to be the constant $0$ (or $\pi$ or $2\pi$) function.
However, consider what happens if we translate $\phi$ in $\theta$ and concatenate the resulting homotopy $\hat{\varphi}$ with the homotopy $\varphi$ above. Explicitly, consider $\Phi : [0,1] \times [0, 2\pi] \rightarrow \mathbb{R}$ defined by $$\Phi(t,\theta) = \varphi(2t, \theta), \text{ if }0 \le t \le 1/2 \text{, and} $$ $$\Phi(t,\theta) = \varphi(2t-1, \theta-\pi/2), \text{ if }1/2 \le t \le 1.$$ Now, the equivariant points for this family are $$\{ (t, \theta) \in [0,1] \times [0,2\pi] : \Phi(t,\theta) = \Phi(t,\theta + \pi) \} =\big(\{0,1/2,1\}\times[0, 2\pi]\big) \cup\big([0,1/2] \times \{0, \pi, 2\pi\}\big) \cup \big( [1/2,1] \times\{\pi/2, 3\pi/2\}\big), \text{ which is pictured as}$$
It follows that there does not exist any function $\Psi :[0,1] \rightarrow [0,2\pi]$ such that $\Psi(t)$ is an equivariant point for all $t$ and $\Psi$ is continuous at $t=1/2$.
I'll additionally remark that, while this example shows that it's impossible to continuously choose equivariant points in a neighborhood of $t=1/2 \in [0,1]$, it is still possible to continuously choose equivariant points in a "one-sided neighborhood" of $t=1/2$. I expect, however, that even this is impossible in general and that a counter-example might be obtained by concatenating infinitely many $\Phi(t, \theta)$ which are suitably rescaled.