Why selection sort is a stable sort algorithm?
Just run an example. I will use colors to distinguish multiple elements with the same value and a $\mid$ to separate the unsorted part from the sorted part.
With the input
$\mid\color{blue}{2},3,\color{red}{2},1$
the first step will swap $\color{blue}{2}$ with $1$.
$1,\mid3,\color{red}{2},\color{blue}{2}$
The order of the $2$s changed. It continues
$1,\color{red}{2},\mid3,\color{blue}{2}$
$1,\color{red}{2},\color{blue}{2},\mid3$
So, the $2$s ended in a different order as they started.
To avoid this you can insert the smallest element of the unsorted part at the beginning of the unsorted part, instead of swapping it.
So, the first step would be
$1,\mid\color{blue}{2},3,\color{red}{2}$
The next step leaves it as it is.
$1,\color{blue}{2},\mid3,\color{red}{2}$
It continues with
$1,\color{blue}{2},\color{red}{2},\mid3$
To prove that this happens in general, observe that if an element with multiplicity is about to be moved, it must be the first element (from left to right) of that value appearing in the unsorted part. So, when it gets moved to the beginning of the unsorted part it continues appearing before all the elements of the same value appearing in the unsorted part and after all the elements of the same value that could already be in the sorted part.