No, duality and symmetry are not the same thing. Although in many contexts "the dual of" is a symmetric relation, this is not invariably the case (e.g. the dual of the dual of a topological vector space need not be the original).
Moreover symmetry is not just about symmetric relations; it has to do mainly with automorphisms of algebraic, geometric or combinatorial structures. Those structure preserving automorphisms (including trivial the identity mapping) form a group, and we'd refer to it as the symmetry group of the structure.
As you note, there are many kinds of symmetry. Some symmetries have order two but many do not. Indeed the group of symmetries may combine elements that have finite order with those having infinite order, elements that have discrete action with some that are continuous mappings. The symmetries of a right circular cylinder, for example, would include discrete actions like reflection in a midplane as well as continuous actions of rotation about the axis.
If you are looking for a fundamental difference, perhaps it should be noted that duality often involves different categories, i.e. the dual may belong to a different category than the original, while symmetry involves not only the same category but actually a mapping of the same object to itself.
Yes, it is true completely independently of the basis for $V$ or the size of our spanning subset $S$. All you need to do is check the axioms for subspace (or axioms for vector space and show containment). Remember Span$(S)$ is all elements of the form $k_{1}s_{1} + ... k_{n}s_{n}$, where each $k_i\in k$ and each $s_i\in S$. Apparently you are using a different definition of Span, but it is quite easy to show the two are equivalent so I will leave that to you (Hint:they are both the smallest subspace containing S).
$1$. If $x , y\in \text {Span} (S)$, then $x+y \in \text {Span} (S)$
Since $x\in \text {Span} (S)$ then $x=k_{x,1}s_{x,1} + ... k_{x,n}s_{x,n}$ for some $k_{x,i}\in k$ and $s_{x,i}\in S$.
Similarly $y=k_{x,1}s_{x,1} + ... k_{x,n}s_{x,n}$, so $x+y=k_{x,1}s_{x,1} + ... k_{x,n}s_{x,n}+k_{x,1}s_{x,1} + ... k_{x,n}s_{x,n}$ so $x+y\in \text{Span} (S)$.
The other axioms are just as straight forward to show so I will leave them to you.
Best Answer
"Span" is less ambiguous and "generate" is more general.
In the contexts of vectors in a vector space, "generated by" and "spanned by" mean the same thing. However, "generate" can mean various other things (for example generation as a module, as an algebra, as a field, etc.) if there is extra structure on the vector space.
A general definition of generation is as follows: let $e_i$ be elements of some structured set $A$. A subset $B \subset A$ is said to be the structure generated by the $e_i$ if it is the intersection of all substructures of $A$ containing the $e_i$. (The key word here is "substructure": the notion of generation changes depending on what kind of structure one is considering on $A$.) Of course this definition only makes sense if the intersection of substructures of $A$ is another substructure, but this is in my experience true of structures to which the word "generate" are applied.