$Rank(A)\leq r \iff$ There are B and C such that $A=BC$ Let $A\in M_{m \times n}$. Then

$\operatorname{rank}(A)\leq r$ $\iff$ there are $B\in M_{m \times r}$ and $C\in M_{r \times n}$ such that $A=BC$.
($\Leftarrow$) $A=BC\Rightarrow A=\sum\limits_{i=1}^r\sum\limits_{j=1}^n C^{ij}B^i\Rightarrow \operatorname{rank}(A)\leq r$
($\Rightarrow$) $\operatorname{rank}(A)\leq r\Rightarrow A_{m \times n}=B_{m \times i}C_{i \times n}=\sum\limits_{i=1}^r\sum\limits_{j=1}^n C^{ij}B^i\Rightarrow$ there are $B\in M_{m \times r}$ and $C\in M_{r \times n}$ such that $A=BC$.
Is it a valid proof? can it be $\iff$ proof? is there an easier proof? 

No, the arguments you're using are not valid.
Suppose $A=BC$, where $B\in M_{m\times r}$ and $C\in M_{r\times n}$. Then the rank of $B$ is at most $r$, so also the rank of $A$ is at most $r$, because $\operatorname{rank}(XY)\le\operatorname{rank}(X)$.
If $\operatorname{rank}(A)=k\le r$, then a full rank decomposition of $A$ is of the form $A=XY$, with $X\in M_{m\times k}$ and $Y\in M_{k\times n}$.
Consider the matrix $B$ obtained from $X$ by adding $r-k$ zero columns at the right and $C$ obtained by $Y$ by adding $r-k$ zero rows at the bottom. Then $B\in M_{m\times r}$, $C\in M_{r\times n}$ and $BC=XY$.
You get a full rank decomposition of $A$ in various ways. For instance the LU-decomposition or the QR-decomposition.
