Prove that convolution with a separable filter is equivalent to convolution on each 1D filter
It requires a good deal of interpretation and a few clear sketches.I will try to provide a clean and easy approach, may be not very rigorous. 

Let $x[n,m]$ be the original image (2D sequence) where $n$ is along the horizontal and $m$ is along the vertical axes and the origin being at the bottom left.

Consider the separable 2D filter sequence $h[n,m] = f[n]g[m]$ where $f[n]$ and $g[m]$ are the corresponding 1D filter sequences along the horizontal and vertical axes respectively.

The 2D output sequence of the convolution is denoted similarly as $$y[n,m] = x[n,m]\star h[n,m]$$ The procedure proceeds by writing the convolutiom sum:

$$ \begin{align}
y[n,m] &amp;= \sum_k \sum_r h[n-k,m-r] x[k,r] \\
&amp;= \sum_k  \sum_r f[n-k] g[m-r] x[k,r] \\
&amp;= \sum_k  f[n-k] \left( \sum_r g[m-r] x[k,r] \right) \\
\end{align}
$$

At the this point, recognize the following: The summation inside the brackets indicate a 1D convolution between the filter $g[m]$ and the $k-th$ column of the original image: $x[k,m]$. Remember that during the summation the dummy index $k$ is fixed, whereas the other dummy index $r$ is ranging through the vertical samples of $x[k,r]$. We shall denote this intermediate sum as a new 2D sequence called $w[k,m]$, parametrized according to $k$ and $m$. That's the intermediate image defined for each value of $k$ and $m$ and formed by convolving each column-$k$ of $x[n,m]$ by the filter sequence $g[m]$. Then the equation follows as: 

$$ \begin{align}
y[n,m]  &amp;= \sum_k  f[n-k] \left( \sum_r g[m-r] x[k,r] \right) \\
&amp;= \sum_k  f[n-k] w[k,m] \\
y[n,m] &amp;= f[n] \star w[n,m] \\
\end{align}
$$

Where we interpret the last line as a 1D convolution between the rows of the intermediate image (for each fixed $m$) and the 1D sequence $f[n]$ along horizontal direction. Which finishes the proof.

Indeed, it's hard to visualize the process, as it's a two step approach. Nevertheless, relience on the assumed interpretations and verificaton by matlab coding would rather help.