There are some explanations for the famous Borsuk-Ulam theorem, e.g. Intuition behind Borsuk-Ulam Theorem, but it seems to me that there are flaws.
Here is my possible remedy for the intuitive proof above:
Let's consider continuous maps $f_1:S^2 \rightarrow \mathbb{R}$ and $f_2:S^2 \rightarrow \mathbb{R}$. Pick any point $x_0 \in S^2$ and choose any path $\alpha \subset S^2$ from $x_0$ to antipodal point $-x_0$. As we move continuously from $x_0$ to $-x_0$ via $\alpha$, then the value $f_1(x_0)-f_1(-x_0)$ turns into a value $-(f_1(x_0)-f_1(-x_0))$. Therefore, in the spirit of intermediate value theorem, there must be at least one one point $z \in S^2$ so that $f_1(z)-f_1(-z)=0$.
Now, let the points $x_0$ and $-x_0$ be fixed and turn the path $\alpha$ one revolution around the sphere $S^2$. It's clear that some point $z$ also moves continuously and forms a closed curve on the sphere (just like in the case of intermediate value theorem when we move y-axis, then at least one one root point also moves continuously all the time). Therefore we have a closed curve $\beta_{x_0} \subset S^2$ that contains only points that satisfy $f_1(z)-f_1(-z)=0$. In the same manner we infer that there is a closed curve $\gamma_{x_0} \subset S^2$ for the function $f_2$.
For any point $p\in \alpha$ we have closed curves $\beta_{p},\gamma_{p}, \subset S^2$. Note that $\beta_{-p}=-\beta_{p}$ and $\gamma_{-p}=-\gamma_{p}$, where the minus sign means that the curve is antipodal in $S^2$. Now suppose that the previous curves don't intersect for any $p\in \alpha$, and that means that e.g. the initial curve $\beta_{x_0}$ is "inside" the initial curve $\gamma_{x_0}$ in sense of fixed direction in $S^2$.
Now we let point $p$ move continuously from $x_0$ to $-x_0$ along path $\alpha$. In that process curve $\beta_{x_0}$ turns in a continuous fashion into a curve $-\beta_{x_0}$, and same with $\gamma_{x_0}$ that turns into $-\gamma_{x_0}$. But now the curve $-\beta_{x_0}$ is "outside" for curve $-\gamma_{x_0}$, so the curves must have crossed during that process. Therefore there is at least one point $s \in S^2$ that satisfy the condition $(f_1(s),f_2(s))=(f_1(-s),f_2(-s))$.
Is this a correct geometric idea for the intuitive proof or are there some pitfalls?
Your 'obvious' statement that a point satisfying $f(x) = f(-x)$ traces out a continuous curve does not seem too obvious - this, to me, is the hardest step of the proof, compared to which the rest is fairly trivial.
In fact, if you can show this, then the last two paragraphs seem unnecessary - simply apply the argument that any closed curve has a pair of antipodal points where the second function is equal to the curve $\beta_{x_0}$