Why does a basis of a vector space need to have the same number of vectors as the vector space dimension? As an example, in $\mathbb{R^3}$, the canonical basis is:\begin{align}
    B &= \left(\begin{bmatrix}
           1 \\
           0 \\
           0
         \end{bmatrix},\begin{bmatrix}
           0 \\
           1 \\
           0
         \end{bmatrix}, \begin{bmatrix}
           0 \\
           0 \\
           1
         \end{bmatrix}\right)
  \end{align} 
So it is a set of 3 vectors. 
There are also different bases in $\mathbb{R^3}$ but they always are sets of 3 vectors. I am wondering why couldn't just \begin{align}
    A &= \begin{bmatrix}
           1 \\
           1 \\
           1
         \end{bmatrix}
  \end{align} 
be considered a basis? Couldn't vector A, multiplied with any scalar reproduce the whole vector space $\mathbb{R^3}$? 

A basis is just any set of linearly independent vectors. Since also $A$ is a (one element) set of linearly independent vectors, it is in fact a basis. But it is neither the basis of $R^3$ nor is it the basis of $R^1$ as you probably might think. It is a basis of a one-dimensional subspace of $R^3$.
Since $R^3$ is defined as the set of all triples $(x,y,z)^T$ with $x,y,z\in R$, it is trivial to see that $R^3$ is just the set of all linear combinations of the basis vectors from $B$:
$$\left( \begin{array}{c} x\\ y\\ z\\ \end{array}\right)=x\cdot\left( \begin{array}{c} 1\\ 0\\ 0\\ \end{array}\right)+y\cdot\left( \begin{array}{c} 0\\ 1\\ 0\\ \end{array}\right)+z\cdot\left( \begin{array}{c} 0\\ 0\\ 1\\ \end{array}\right)$$
Now suppose that $A$ was also a basis of $R^3$. Then it would have to be possible to write all the vectors from $R^3$ as linear combinations of vectors from $A$. Since $B$ also contains vectors from $R^3$, these would all need to be expressible by the 'basis' A. Hence you would need to be able to write for example
$$\left( \begin{array}{c} 1\\ 0\\ 0\\ \end{array}\right)=\lambda \left( \begin{array}{c} 1\\ 1\\ 1\\ \end{array}\right)$$
and the same for the other basis vectors from $B$.
Splitting this equation into components leads to equations for $\lambda$, namely
$$\lambda=1$$
and
$$\lambda=0$$
which is an obvious logical contradiction. Hence, $A$ cannot be a basis of $R^3$.
You can extend this line of thought to arbitrary finite dimension, where you conclude that one basis must be expressible by another basis and where you can show that bases with different number of basis vectors lead to a contradiction to the assumption, that the basis vectors are linearly independent (which must always be the case for a basis). The final conclusion of this proof will be that if there is a finite basis for a vector space, every other basis of the same space must have the same number of basis vectors. Then you call this number the 'dimension' of that vector space. 
By the way, there are also infinite-dimensional vector spaces, where things are complicated by the fact, that in case of a countable basis you can number elements (basis vectors) in different ways, and there are even uncountable sets (the real numbers are an example of a uncountable set, although it is of course not a basis of anything) which cannot by numbered at all.
