Is a spanning set not just a linearly independent set of vectors, or is there a difference?
The context of this question being the following theorem:
"In any vector space, if $|I|$ is a linearly independent set and $|S|$ is a spanning set, then $|I| \leq |S|$"
On
$\{(1,0),(0,1),(1,1)\}$ is a spanning set of the $\mathbb{R}$-vector space $\mathbb{R}^2$ but is not linearly independent over $\mathbb{R}$.
Let $k$ be a field and $V$ be a $k$-vector space. Let $S\subset V$ be a finite set, $S$ is said to be a spanning set of $V$ if and only if each element of $V$ can be express as a finite linearly combination of elements in $S$.
On
A set of vectors can be spanning without being linearly independent and vice versa. Compare the definitions:
A set of vectors $\{v_1,\ldots,v_n\}$ spans a vector space $V$ if every vector $x \in V$ is a linear combination of $v_1,\ldots,v_n$.
For example, the set of vectors $\{(1,0), (0,1)\}$ spans $\mathbb{R}^2$, as does the set of vectors $\{(1, 0), (0, 1), (\pi, \sqrt{2})\}$.
A set of vectors $\{v_1, \ldots, v_n\}$ is linearly independent if whenever $\alpha_1 v_1 + \cdots + \alpha_n v_n = 0$, it must be the case that $\alpha_1 = \cdots = \alpha_n = 0$.
For example, the set of vectors $\{(1,0), (0,1)\}$ is linearly independent in $\mathbb{R}^2$, as is the set of vectors $\{(1, 0)\}$.
However, the spanning set $\{(1, 0), (0, 1), (\pi, \sqrt{2})\}$ is not linearly independent, since $$ -\pi(1,0) - \sqrt{2} (0, 1) + (\pi, \sqrt{2}) = (0, 0), $$ but $-\pi \neq 0$ and $-\sqrt{2} \neq 0$. Moreover, the linearly independent set $\{(1, 0\}$ does not span $\mathbb{R}^2$ since, for example, $(0,1)$ is not a scalar multiple of $(1,0)$.
On
A spanning set is a set of vectors who when combined in all possible combinations cover the space equal to some $R^n$.
However, many of these vectors might just be scalar multiples of each other. As a result, many are redundant and serve no purpose towards covering $R^n$. If you remove all the ones that can be expressed as some linear combination of the others, you are left with a "linearly independent" set of vectors that still span over all of $R^n$.
For example, in $\mathbb R^3$ the set $$\{ (1,0,0), (0,1,1) \}$$ is linearly independent meaning no member of the set can be written as a linear combination of other members of that set. But linear combinations of those vectors clearly don't span $\mathbb R^3$.