$Rank(A)\leq r \iff$ 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? 
 A: 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.
A: It is not a valid proof, because as far as I can see you are just writing formulas that express matrix multiplication, and any reasoning that might be going on is completely implicit (which is polite for: absent).
A correct reasoning is much easier if you abstract away from the concrete matrix representation, which is just leads to distraction by the nitty-gritty details of matrix multiplication. Interpret matrices as corresponding to linear maps between finite dimensional vector spaces with respect to certain bases; lacking any specific context you can take these spaces to be of the form $K^n$ (where $K$ is you base field) equipped with their standard bases, but the argument will be independent of this particular choice. The rank of a matrix (the dimension of the span of the columns) is the rank (dimension of the image) of the corresponding linear map, because the columns are just the coordinates on the basis at arrival of a spanning set of the image, and the coordinate map is an isomorphism (so preserves dimensions of subspaces).
So now you are asking why the rank of a linear map $a:U\to W$ is at most$~r$ if and only if it can be written as the composition $a=b\circ c$ of linear maps $c:U\to V$ and $b:V\to W$ where $\dim V=r$. I don't see a proof that does both implications at once, but both of them are easy. Supposing one has such a decomposition $a=b\circ c$, then the image $a(U)=b(c(U))$ is contained in the image $b(V)$ (since $c(U)\subseteq V$), which has dimension $\def\rk{\operatorname{rk}}\rk(b)=\dim(V)-\dim(\ker(b))$ by rank nullity; then $$\rk(a)=\dim(a(U))\leq\dim(b(V))=\rk(b)=\dim(V)-\dim(\ker(b))\leq\dim(V)=r.$$
Conversely suppose $\rk(a)\leq r$. One can always factor $a$ through its own image $a:U\to a(U)\hookrightarrow W$, so in case $\rk(a)=r$ one can simply take $V=U(A)$, with $c$ the map $a$ itself but made surjective by replacing the space at arrival with the actual image $u(A)$, and $b$ the inclusion of that image into $W$. If $\rk(a)<r$ one needs to add $d=r-\rk(a)$ to the dimension of$~a(U)$ in forming$~V$, which can be done by taking $V=a(U)\times K^d$; now $c$ can be taken to be the composition $U\to a(U)\hookrightarrow a(U)\times K^d$, and $b$ the composition, with as first map projection on the first factor: $a(U)\times K^d\to a(U)\hookrightarrow W$.
A: Here is a simple proof: let $u\colon \mathbf R^n\rightarrow \mathbf R^m$ be the linear map with matrix $A$ in the canonical basis. Note that $\operatorname{rank} A <r$  if and only if there exist  linear maps $f\colon\mathbf R^n\rightarrow \mathbf R^r $ and $g\colon\mathbf R^r\rightarrow \mathbf R^m$ such that the following diagram commutes:

i.e. such that $u=g\circ f$.
Indeed, if  $\operatorname{rank} A=\operatorname{rank} u =s\leq r$ there exists an isomorphism $\varphi\colon\operatorname{Im} u \simeq \mathbf R^s$. Consider the linear map $\bar u\colon \mathbf R^n\rightarrow \operatorname{Im}u, x\mapsto u(x)$, the  injection $i\colon\mathbf R^ s\hookrightarrow \mathbf R^r, \enspace x\mapsto (x,0,\dots,0)$ ($r-s$ zeroes) and set $f=i\circ \varphi\circ\bar u$.
Now for $g$: let $j\colon\operatorname{Im} u \hookrightarrow \mathbf R^m$ be the canonical injection and $p\colon \mathbf R^r\rightarrow\mathbf R^ s$ the projection on the first $r$ factors. Set $g=j\circ \varphi^{-1}\circ p$. We have:
$$ g\circ f=(j\circ\varphi^{-1}\circ p)\circ(i\circ \varphi\circ\bar u)=j\circ(\varphi^{-1}\circ (p\circ i)\circ \varphi)\circ\bar u=j\circ\varphi^{-1}\circ \varphi\circ\bar u=j\circ u=u. $$
Conversely, if $u=g\circ f$ as indicated, $\operatorname{rank} u \le\operatorname{rank} g\le r$.
Now if  $B\in\mathcal M_{m\times r}$ and $C\in\mathcal M_{r\times n}$ are the matrices associated with $g$ and $f$ respectively, the decomposition  $u=g\circ f$ means $A=BC$.
