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This is the corrected proof of the satement in the case when $F'$ is supposed to be continuous. We also assume that $F'$ never vanishes, othevise the statement is obviuous. The comments below are about previous versions of my proof, which were not correct. So the comments are not relevant anymore to this answer (sorry for that).

I will use the reasoning of Ilya (please read his answer. His idea is that we conisder the curve $\frac{F'}{|F'|}$ in the unit $S^{n-1}$). Then we need to prove a little lemma.

Lemma. Consider a continous curve $C$ on a sphere $S^{n-1}$ that doesn't lie inside any half-sphere. Prove that there exists an equator that intersects the curve in at least $n$ points.

Proof. The center of the sphere belongs to the convex hull of $C$ by the assumption of the lemma. This means that there are $n+1$ or less points on $C$ such that the simplex with vertices in these points contains the centre $0$ of the sphere. If the number of points is less than $n+1$, we are done. Suppose that the number of points is in fact $n+1$ and they are linearly inependent. We call the points $x(t_1),...,x(t_{n+1})$ where $t_i$ are odered for. Consider now the hyperplane $H$ generated by points $x(t_1),...,x(t_{n-1})$ and the center of the sphere. $H$ cuts $S^{n-1}$ in two halves. Since by constructions the convex hull of $x(t_i)$ contains $0$, the points $x(t_n), x(t_{n+1})$ are in different half spaces with respect to the plane. So this plane intersects the part of curve C contained between $t_n$, $t_{n+1}$ in some point $x(t_n')$. The lemma is proved.

This is the corrected proof of the satement in the case when $F'$ is supposed to be continuous. We also assume that $F'$ never vanishes, othevise the statement is obviuous. The first four comments below are about previous versions of my proof, which were not correct. So first 4 comments are not relevant anymore to this answer (really sorry for that).

I will use slightly modified reasoning of Ilya (please read his answer. His idea is that we conisder the curve $\frac{F'}{|F'|}$ in the unit $S^{n-1}$). Then we need to prove a little lemma.

Lemma. Consider a continous curve $C$ on a sphere $S^{n-1}$ that doesn't lie inside any half-sphere parametrised by an inteval $[0,1]$. Prove that there exists an equator that intersects the curve for at least $n$ different values of the parameter.

Proof. The center of the sphere belongs to the convex hull of $C$ by the assumption of the lemma. This means that there are $n+1$ or less points on $C$ such that the simplex with vertices in these points contains the centre $0$ of the sphere. If the number of points is less than $n+1$, we are done. Suppose that the number of points is in fact $n+1$ and they are linearly inependent. We call the points $x(t_1),...,x(t_{n+1})$ where $t_i$ are odered for. Consider now the hyperplane $H$ generated by points $x(t_1),...,x(t_{n-1})$ and the center of the sphere. $H$ cuts $S^{n-1}$ in two halves. Since by constructions the convex hull of $x(t_i)$ contains $0$, the points $x(t_n), x(t_{n+1})$ are in different half spaces with respect to the plane. So this plane intersects the part of curve C contained between $t_n$, $t_{n+1}$ in some point $x(t_n')$. The lemma is proved.

This is the corrected proof of the satement in the case when $F'$ is supposed to be continuous. We also assume that $F'$ never vanishes, othevise the statement is obviuous.

I will use the reasoning of Ilya (please read his answer. His idea is that we conisder the curve $\frac{F'}{|F'|}$ in the unit $S^{n-1}$). Then we need to prove a little lemma.

Lemma. Consider a continous curve $C$ on a sphere $S^{n-1}$ that doesn't lie inside any half-sphere. Prove that there exists an equator that intersects the curve in at least $n$ points.

Proof. The center of the sphere belongs to the convex hull of $C$ by the assumption of the lemma. This means that there are $n+1$ or less points on $C$ such that the simplex with vertices in these points contains the centre $0$ of the sphere. If the number of points is less than $n+1$, we are done. Suppose that the number of points is in fact $n+1$ and they are linearly inependent. We call the points $x(t_1),...,x(t_{n+1})$ where $t_i$ are odered for. Consider now the hyperplane $H$ generated by points $x(t_1),...,x(t_{n-1})$ and the center of the sphere. $H$ cuts $S^{n-1}$ in two halves. Since by constructions the convex hull of $x(t_i)$ contains $0$, the points $x(t_n), x(t_{n+1})$ are in different half spaces with respect to the plane. So this plane intersects the part of curve C contained between $t_n$, $t_{n+1}$ in some point $x(t_n')$. The lemma is proved.

This is the corrected proof of the satement in the case when $F'$ is supposed to be continuous. We also assume that $F'$ never vanishes, othevise the statement is obviuous. The comments below are about previous versions of my proof, which were not correct. So the comments are not relevant anymore to this answer (sorry for that).

I will use the reasoning of Ilya (please read his answer. His idea is that we conisder the curve $\frac{F'}{|F'|}$ in the unit $S^{n-1}$). Then we need to prove a little lemma.

Lemma. Consider a continous curve $C$ on a sphere $S^{n-1}$ that doesn't lie inside any half-sphere. Prove that there exists an equator that intersects the curve in at least $n$ points.

Proof. The center of the sphere belongs to the convex hull of $C$ by the assumption of the lemma. This means that there are $n+1$ or less points on $C$ such that the simplex with vertices in these points contains the centre $0$ of the sphere. If the number of points is less than $n+1$, we are done. Suppose that the number of points is in fact $n+1$ and they are linearly inependent. We call the points $x(t_1),...,x(t_{n+1})$ where $t_i$ are odered for. Consider now the hyperplane $H$ generated by points $x(t_1),...,x(t_{n-1})$ and the center of the sphere. $H$ cuts $S^{n-1}$ in two halves. Since by constructions the convex hull of $x(t_i)$ contains $0$, the points $x(t_n), x(t_{n+1})$ are in different half spaces with respect to the plane. So this plane intersects the part of curve C contained between $t_n$, $t_{n+1}$ in some point $x(t_n')$. The lemma is proved.

This is the proof of the satement in the case when $F'$ is supposed to be continuous. If you down vote it, please explain what is wrong here.

I will use the reasoning of Ilya (please read his answer. His idea is that we conisder the curve $\frac{F'}{|F'|}$ in the unit $S^{n-1}$). We need to prove a little lemma.

Lemma. Consider a continous curve $C$ on a sphere $S^{n-1}$ that doesn't lie inside any half-sphere. Prove that there exists an equator that intersects the curve in at least $n$ points.

Proof. The center of the sphere belongs to the convex hull of $C$ by the assumption of the lemma. This means that there are $n+1$ or less points on $C$ such that the simplex with vertices in these points contains the centre $0$ of the sphere. If the number of points is less than $n+1$, we are done. Suppose that the number of points is in fact $n+1$ and they are linearly inependent. We call the points $x(t_1),...,x(t_{n+1})$ where $t_i$<$t_j$ for $i$<$j$. Consider now the hyperplane $H$ generated by points $x(t_1),...,x(t_{n-1})$ and the center of the sphere. $H$ cuts $S^{n-1}$ in two halves. Since by constructions the convex hull of $x(t_i)$ contains $0$, the points $x(t_n), x(t_{n+1})$ are in different half spaces with respect to the plane. So this plane intersects the part of curve C contained between $t_n$, $t_{n+1}$. We are done.

This is the corrected proof of the satement in the case when $F'$ is supposed to be continuous. We also assume that $F'$ never vanishes, othevise the statement is obviuous.

I will use the reasoning of Ilya (please read his answer. His idea is that we conisder the curve $\frac{F'}{|F'|}$ in the unit $S^{n-1}$). Then we need to prove a little lemma.

Lemma. Consider a continous curve $C$ on a sphere $S^{n-1}$ that doesn't lie inside any half-sphere. Prove that there exists an equator that intersects the curve in at least $n$ points.

Proof. The center of the sphere belongs to the convex hull of $C$ by the assumption of the lemma. This means that there are $n+1$ or less points on $C$ such that the simplex with vertices in these points contains the centre $0$ of the sphere. If the number of points is less than $n+1$, we are done. Suppose that the number of points is in fact $n+1$ and they are linearly inependent. We call the points $x(t_1),...,x(t_{n+1})$ where $t_i$ are odered for. Consider now the hyperplane $H$ generated by points $x(t_1),...,x(t_{n-1})$ and the center of the sphere. $H$ cuts $S^{n-1}$ in two halves. Since by constructions the convex hull of $x(t_i)$ contains $0$, the points $x(t_n), x(t_{n+1})$ are in different half spaces with respect to the plane. So this plane intersects the part of curve C contained between $t_n$, $t_{n+1}$ in some point $x(t_n')$. The lemma is proved.