2-Sphere is connected I am supposed to show, that if I have a continuous function $F:S^2→ (\mathbb{R}, | · |)$, with $S^2$ being the 2-Sphere in $\mathbb{R^3}$, that there is a point  where $F(x)=F(-x)$. Now I know how to do a majority of the proof. One has to show that $S^2$ is connected, therefore the image of $G(x):=F(x)-F(-x)$ is connected and because in $\mathbb{R}$ a set is connected iff it is an interval, and because $G(-x)=-G(x)$ we know that the image of $G(x)$ has to contain $0$, and therefor the function $F(x)$ has a zero point. What I am struggling with however is showing that $S^2$ is connected. I do not have any idea how to show that, at the moment i am trying to show that it is path connected, but I am having trouble defining a function that alloqws me to prove this. Any ideas?
Edit: We have not gone into particularly much depth on the topic, as I am only a second semester math student so I would be happy if someone could give a simple solution. Thanks!

Do you know that union of connected sets with a point in common is connected? If so, then the sphere is the union of two disks along their boundary, and disks are connected as you may well know.
Another simple solution is parametrizing the sphere, which takes the closed square surjectively in the sphere. Since continuous maps take connected sets to connected sets, the result follows.

If you know the concept of path-connectedness, you can join two points in a sphere in the following way (and note this generalizes to $S^n$) :
For any two non-antipodal points $x,y$, join them by a line in $x+t(y-x)$ in $\mathbb{R}^n%$ and consider the path
$$\frac{1}{\Vert x+t(y-x)\Vert}(x+t(y-x)).$$
Therefore, we can connect any two non-antipodal points on the sphere. If they are antipodal points, take any other $z$ and concatenate the paths.
Exercise: Why is it important that $x,y$ are not antipodal points?
