How to prove and interpret $\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B))$? 
Let $A$ and $B$ be two matrices which can be multiplied. Then $$\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B)).$$

I proved $\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ by interpreting $AB$ as a composition of linear maps, observing that $\operatorname{ker}(B) \subseteq \operatorname{ker}(AB)$ and using the kernel-image dimension formula. This also provides, in my opinion, a nice interpretation: if non-stable, under subsequent compositions the kernel can only get bigger, and the image can only get smaller, in a sort of loss of information.
How do you manage $\operatorname{rank}(AB) \leq \operatorname{rank}(A)$? Is there a nice interpretation like the previous one?
 A: Here is another simple answer. When you multiply a matrix and a vector $Ax$ you end up with a linear combination of the columns of $A$.
$$ Ax = \; x_1\,A_1 \;+\; x_2\,A_2 \;+\; x_3\,A_3 \;+\;\; ...\;\; \\ $$
When we multiply two matrices $AB = C$, we have $AB_i = C_i$, which means that each column of $C$ is a linear combination of the columns of $A$, so $\text{rank}(AB) \leq \text{rank}(A)$. To show that $\text{rank}(AB) \leq \text{rank}(B)$ we follow a similar argument -- when you multiply $x^{\top}B$, you end up with a linear combination of the rows of $B$.
A: Yes. If you think of A and B as linear maps, then the domain of A is certainly at least as big as the image of B. Thus when we apply A to either of these things, we should get "more stuff" in the former case, as the former is bigger than the latter.
A: Already you have  proved $$rank(AB)\leq rank(B)$$
For other part 
rank of $ A=$ dim range $A$
As range $AB \subset $ range $A\implies $ dim range $AB\leq $ dim range $A$ . Hence $$rank(AB)\leq rank (A)$$
A: Once you have proved $\operatorname{rank}(AB) \le \operatorname{rank}(A)$, you can obtain the other inequality by using transposition and the fact that it doesn't change the rank (see e.g. this question). 
Specifically, letting $C=A^T$ and $D=B^T$, we have that $\operatorname{rank}(DC) \le \operatorname{rank}(D) \implies \operatorname{rank}(C^TD^T)\le \operatorname{rank} (D^T)$, which is $\operatorname{rank}(AB) \le \operatorname{rank}(B)$.
A: Let $ m \le n, A \in M_{m\times n}, B\in M_{n\times m}$.
$\mbox{rank } A\le m$ and $\text{rank }B\le m$. (Obvious fact as rank A = dimension of the columnspace of A = dimension of the row space of A.)
Let $E_{n\times n}B$ be the row echelon form of $B$ and let $AE_{m\times m} $ be the column echelon form of $A$. ($E_{n\times n} ,E_{m\times m}$ are elementary matrices.)
We know $\operatorname{rank}(BA)=\operatorname{rank}(E_{n\times n}BA )=\operatorname{rank}(E_{n\times n}BAE_{m\times m} )$. 
But $E_{n\times n}BAE_{m\times m} =\begin{pmatrix}
L&0\\
0&0\\
\end{pmatrix},$ where $L$ is an $k\times l$ matrix with $k\le \operatorname{rank}(B),l\le \operatorname{rank}(A)$.
So $\operatorname{rank}(E_{n\times n}BAE_{m\times m} )=\operatorname{rank}\begin{pmatrix}
L&0\\
0&0\\
\end{pmatrix}\le \min\{k,l\}\le \min\{\mbox{rank } A,\mbox{rank }B\}.$
A: Another way of showing that 
$\text{Rank} (AB) \leq \text{Rank}(B)$ without using rank-nullity:
Note that if $v_1,\dots,v_n$ is a basis of $\text{Range} B$, then $Av_1,\dots,Av_n$ is a spanning list of $\text{Range} AB$. 
A: Note that $\text{Col}(AB) ⊆ \text{Col}(A)$ since given $y ∈ \text{Col}(AB)$ we can choose $x ∈ F$ and then we have $y = (AB)x = A(Bx) ∈ \text{Col}(A)$. 
Since $\text{Col}(AB) ⊆ \text{Col}(A)$, any basis for $\text{Col}(AB)$ can be extended to a basis for $\text{Col}(A)$ and so $\dim \text{Col}(AB) ≤ \dim \text{Col}(A)$, that is $\text{rk}(AB) ≤ \text{rk}(A).$
Note that $\text{Null}(B) ⊆ \text{Null}(AB)$ since given $x ∈ \text{Null}(B)$ we have $(AB)x = A(Bx) = A 0 = 0$ so that $x ∈ \text{Null}(AB)$. 
Since $\text{Null}(B) ⊆ \text{Null}(AB)$, as above we have $\dim \text{Null}(B) ≤ \dim \text{Null}(AB)$, that is, $\text{Nullity}(B) ≤\text{Nullity}(AB)$. Thus $\text{rk}(AB) = n − \text{Nullity}(AB) ≤ n − \text{Nullity}(B) = \text{rk}(B).$
A: Prove first that if $f:X\to Y$ and $g:Y\to Z$ are functions between finite sets, then 
$|g(f(X))| \leq \min \{ |f(X)|, |g(Y)| \}.$
Then use the same idea.
A: Each column of $AB$ is a linear combination of columns of $A$ so $\text{Range}(AB)\subseteq \text{Range} (A)$ equivalently $rank(AB)\leq rank(A).$
Similarly,
Each row of $AB$ is a linear combination of rows of  $B$ so $rowspace(AB)\subseteq rowspace(B)$.
Since rowspace=columnspace.
Thus $rank(AB)\leq rank(B)$.
From both of the inequality we deduce that $rank(AB)\leq \min\{rank A, rank B\}$.
