Take, for example $V = \mathbb R ^2$, the $x$-$y$ plane. Write the vectors as coordinates, like $(3,4)$.
Such a coordinate could be written as a sum of its $x$ component and $y$ component: $$(3,4) = (3,0) + (0,4)$$
and it could be decomposed even further and written in terms of a "unit" x vector and a "unit" y vector:
$$(3,4) = 3\cdot(1,0) + 4\cdot(0,1).$$
The pair $\{(1,0),(0,1)\}$ of vectors span $\mathbb R^2$ because ANY vector can be decomposed this way:
$$(a,b) = a(1,0) + b(0,1)$$
or equivalently, the expressions of the form $a(1,0) + b(0,1)$ fill the space $\mathbb R^2$.
It turns out the $(1,0)$ and $(0,1)$ are not the only vectors for which this is true. For example if we take $(1,1)$ and $(0,1)$ we can still write any vector:
$$(3,4) = 3\cdot(1,1) + 1\cdot(0,1)$$
and more generally
$$(a,b) = a \cdot (1,1) + (b-a)\cdot(0,1).$$
This fact is intimately linked to the matrix $$\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$$ whose row space and column space are both two dimensional.
I would keep going, but your question is general enough that I could write down an entire linear algebra course. Hopefully this gets you started.
Well then, let us pick any point in $\mathbb R^2$, e.g. the point $(a,b)$.
Let us define $x = a$ and $y=\frac{b-2a}{3}$ then you can easily see that:
$$
\begin{align*}
(a,[b-2a]/3) = (x,y) \\
\iff (a,b-2a) = (x,3y) \\
\iff (a,b) = (x,2a+3y) \\
\iff (a,b) = (x,2x+3y)
\end{align*}
$$
So $(a,b)$ really is in the span of $\{(1,2),(0,3)\}$.
As we can choose any $(a,b) \in \mathbb R^2$, we know that those vectors span the whole $\mathbb R^2$.
Best Answer
Firstly, assume for contradiction that $B \setminus \{ \mathbf{b}_{n} \} = \{ \mathbf{b}_{1}, \ldots, \mathbf{b}_{n - 1} \}$ spans $V$. Then there exists $c_{1}, \ldots, c_{n - 1} \in \mathbb{R}$ such that $\mathbf{b}_{n} = c_{1} \mathbf{b}_{1} + \cdots + c_{n - 1} \mathbf{b}_{n - 1}$. But this would mean that $c_{1} \mathbf{b}_{1} + \cdots + c_{n - 1} \mathbf{b}_{n - 1} + (-1) \mathbf{b}_{n} = \mathbf{0}$, so $B$ is linearly dependent, a contradiction.
Similarly, let $B' = B \cup \{ \mathbf{b}_{n + 1} \} = \{ \mathbf{b}_{1}, \ldots, \mathbf{b}_{n + 1} \}$, where $\mathbf{b}_{n + 1} \in V$. Then since $V = \operatorname{span} (B)$, there exist constants $c_{1}, \ldots, c_{n} \in \mathbb{R}$ such that $c_{1} \mathbf{b}_{1} + \cdots + c_{n} \mathbf{b}_{n} = \mathbf{b}_{n + 1}$. But this would mean that $c_{1} \mathbf{b}_{1} + \cdots + c_{n} \mathbf{b}_{n} + (- 1) \mathbf{b}_{n + 1} = \mathbf{0}$, so $B'$ is dependent.