I know that my question has already an answer here, but I have proved it another way and I want to see whether my proof is true or not?
If we assume $A$ , $B$ and $A+B$ are respectively the matrices of linear transformations $T_A$ , $T_B$ and $T_{A+B}$ with ranges $R_A$ , $R_B$ , $R_{A+B}$ and ranks $rk(A)$ , $rk(B)$ and $rk(A+B)$ then we can have the following proof:
Consider $\{y_1,y_2,\dots, y_{rk(A)}\}$ as a a basis for the range of the matrix $A\,(R_A)$
$$R_A=Span\{y_1,y_2,\dots ,y_{rk(A)}\}$$
and $\{\acute y_1,\acute y_2,\dots,\acute y_{rk(B)}\}$ as a basis for the range of the matrix $B\,(R_B)$
$$R_B=Span\{\acute y_1,\acute y_2,\dots ,\acute y_{rk(A)}\}$$
$\forall \,X\in\mathbb R^n$ we can say $AX\in R_A$ and $BX\in R_B$ so
$\exists\,\alpha_1,\alpha_2,\dots,\alpha_{rk(A)}\in\mathbb R $ such that $AX=\alpha_1y_1+\alpha_2y_2+\dots+\alpha_{rk(A)}y_{rk(A)}$ and
$\exists\,\beta_1,\beta_2,\dots,\beta_{rk(B)}\in\mathbb R $ such that $BX=\beta_1\acute y_1+\beta_2\acute y_2+\dots+\beta_{rk(B)}\acute y_{rk(B)}$ so
$$AX+BX=(A+B)X=\alpha_1y_1+\alpha_2y_2+\dots+\alpha_{rk(A)}y_{rk(A)}+\beta_1\acute y_1+\beta_2\acute y_2+\dots+\beta_{rk(B)}\acute y_{rk(B)}$$
We now that $(A+B)X\in R_{A+B}$ (range of matrix $A+B$) so we can say
$$R_{A+B}=Span\{\underbrace{y_1,y_2,\dots,y_{rk(A)}}_{rk(A)\text{times}},\underbrace{\acute y_1,\acute y_2,\dots, \acute y_{rk(B)}}_{rk(B)\text{times}}\}$$
So we see these $rk(A)+rk(B)$ matrice create the space $R_{A+B}$ but we don't know whether all of them are linearly independent or not. There are two cases:
1- All of $rk(A)+rk(B)$ vectors are linearly independent, so they create a basis for the space $R_{A+B}$ hence $dim(R_{A+B})=rk(A+B)=rk(A)+rk(B)$
2- Some of these vectors are linearly dependent to others, we should omit the linearly dependent vectors in order to find a basis for $R_{A+B}$ hence $dim(R_{A+B})=rk(A+B)\lt rk(A)+rk(B)$
Considering these two cases, we always have:
$$rk(A+B)\le rk(A)+rk(B)$$
