For a matrix to be in Reduced Row Echelon Form it must satisfy the following conditions:
The first non-zero entry in any row is the number $1$. These are called pivots. This implies that every row has a $0$/$1$ pivot. Also, the first non-zero element of any non-zero row appears in the later column (furthest to the right) than the first non-zero element of the preceding row.
The pivot is the only non-zero entry in the column. This implies that each column can have a $0$/$1$ pivot.
The rows are ordered so that any rows consisting of all $0$'s are at the bottom of the matrix, i.e. all non-zero rows precede zero rows.
While the matrix $G$ satisfies condition $1$, since the pivot of each row is the number $1$ and each pivot appears in a later column to the right and also satisfies condition $3$, since the row consisting pure $0$'s is at the bottom (the none-zero rows precede the zero row)...
$G$ however, does not satisfy condition $2$, as both columns $3$ and $4$ have non-zero entries besides their pivots.
The below matrix is $G$ in reduced row echelon form, notice that in all columns with leading ones, the number $1$ (the pivot), is only non-zero entry in the column.
$$G=\left[
\begin{array}
11&0&0&20\\
0 & 1 & 0 & 9 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 0\\
\end{array}
\right]
$$
It is correct but there are some nitpicks to make. For instance:
You write 'Suppose $\mathbf{R} \neq \mathbf{I}$. Then $\mathbf{R}$ must have a leading 1 (call it $x_{i,j}$) which is located in $i$th row and $j$th column and $j>i$.'
Strictly speaking there is a second possibility and that is when $\mathbf{R}$ doesn't contain any non-zero elements at all. Of course this case is easy to handle, but perhaps you should mention it.
Also, depending on your audience, it might not really be clear that (when $x_{i, j}$ exists) we have $j > i$. The fact that $\mathbf{R} \neq \mathbf{I}$ only implies that $j \neq i$ so you might want to elaborate a bit on why $j < i$ is ruled out.
In the converse direction there are two typos: 'if $x_{i,i} = 1$, then $i = j$' should probably read 'if $x_{i, j} = 1$ then $i = j$'. Just following that you write 'because we've shown that if $i > j$ then $\mathbf{R}$ will have row of zeros', but what we have shown is something else, namely that if $i < j$ then $\mathbf{R}$ will have row of zeros.
Of course, also at this point the reader might wonder what happens in the case that $i > j$ (i.e. the case I wrote about earlier). I recommend you include it.
There is something else: you write two proofs now one that not being the identity implies having a row of zeroes and one that not having a row of zeroes implies being the identity matrix. This is nice towards the reader, but since both are logically equivalent (to each other and to the statement of the proposition) strictly speaking you only need one. (This last issue is not a problem with your proof but something that it is good to be aware of)
Best Answer
This is not true, consider matrices with more rows than columns like $$\left(\begin{matrix}1 & 0 \\ 0 & 1 \\ 0 & 0\end{matrix}\right).$$ Free variables are implied by columns without pivot element.
Yes, consider $$\left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0\end{matrix}\right).$$
This depends on the size of the base field. If it is infinite (like $\mathbb{Q}$ or $\mathbb{R}$) then yes. Otherwise no.