Here is a differential geometric proof. To me it is quite intuitive, since it not uses so much topology, despite the fact that the identity on $\Bbb S^1$ is not contractible. I am not totally convinced that this works, but if so, maybe someone has a reference for me?
You could use the stereographic projection where you remove the north pole. For a vector field $F'$ on the sphere there is a corresponding vector field $F$ on $\Bbb R^2$,
The vector field $F$ can be assumed to be normalized. Then for any $r>0$ you get a self map $F_r$ of $\Bbb S^1$ which is the composition
$$ \Bbb S^1 \hookrightarrow \Bbb R^2 \xrightarrow{\cdot r} \Bbb{R}^2 \xrightarrow{F} \Bbb S^1 \subset \Bbb{R}^2.$$
Since $F$ is continuous for sufficiently small $r>0$ the map $F_r$ is contractible.
If we now show that for sufficiently large $r$ this map is homotopic to the identity, this is a contradiction, since the identity on one-sphere is not null homotopic.
In order to do this we have to take a look at the stereographic projection who the tangent vectors are mapped.
The stereographic projection with north pole removed is given by
$$ P_N (x,y,z) = \frac{1}{1-z} (x ,y ) $$
and its Jacobian matrix is
$$
\begin{pmatrix}
\frac{1}{1-z} & 0 & \frac{x}{(1-z)^2} \\
0 & \frac{1}{1-z} & \frac{y}{(1-z)^2} \\
\end{pmatrix}.
$$
Using partitions of unity we can assume that in a small neighborhood of $N$ the vector field $F'$ is given by $F'(x,y,z)=(0,-z,y)$.
Using the Jacobian matrix we know that for large $r$ the corresponding vector field $F$ on $\Bbb R^2$ is given by
$$ F(P_N(x,y,z)) = \left( \frac{xy}{(1-z)^2} , \frac{y^2}{(1-z)^2}- \frac{z}{1-z} \right) . $$
composing with the inverse of the stereographic projection
$$ P_N^{-1}(x,y) = \frac{1}{x^2+y^2} ( 2x, 2y, x^2+y^2-1) $$
we get
$$ F(x,y) = (4xy, 3 y^2 -x^2 +1 ). $$
Switching to to polar coordinates we get
$$ F(\varphi, r) = r^2 \left( 4 \cos \varphi \sin \varphi, 3 \sin^2 \varphi -\cos^2 \varphi + \frac{1}{r^2} \right) = r^2( 2 \sin (2\varphi), 1- 2\cos (2 \varphi) + \frac{1}{r^2} ) . $$
For big $r$ this defines a circle. So $F_r$ has a non zero mapping degree, in fact it is homotopic to the identity.
This completes the proof.
Edit:
Here is some more intuitive argument. Imagine you have a vector field F on S2 ⊂ R3 given on the equator in the x-y-plane by F (x, y, z) = (y, −x, 0). Pictorially this are vectors which ”move” around the circle. Now if you move the plane to the south pole you can look how these vectors rotate. To make this more precise you could project the tangent vectors to the x-y-plane. For a fixed angle in the x-y-plane when you move down the z-axis to the south pole, the projected tangent vector has to rotate into the direction of the tangent vector at south pole. Now what is the problem? The projection maps only zero vectors to zero, since on the x-y-plane the vector field was assumed to be non-vanishing. So if you go down the z-axis, all tangent vectors on points parallel to the x-y-plane will turn to the same direction, but this can not be unless you have a zero or a discontinuity at some point.
It think this is quite intuitive. What is the ”problem” with that? In the prove before I made a local assumption, which is easy to make. But here I made a huge restriction by fixing the vector field on the equator.
Now I claim this is not as bad as it might looks. Here is a recipe how you can turn it into an almost proof. Take a non vanishing tangent field F , then approximate it by an smooth tangent field. Now you need to find one integral curve which gives you a loop on the sphere. Then choose a point inside but not on the curve and take it as south pole. Then with small perturbations ensure that the tangent field does not vanish when projected to the plane though equator. Having this you can proceed as before.
