[tex]\{a,b,c,\ldots,j\}[/tex] has ten elements, so any proper subset has at most nine of these elements.
The number of ways of taking any [tex]0\le n\le 9[/tex] letters from this set is given by the binomial coefficient,
[tex]\dbinom{10}n=\dfrac{10!}{n!(10-n)!}[/tex]
and in particular, the total number of ways to picking proper subsets is
[tex]\displaystyle\sum_{n=0}^9\binom{10}n=\dbinom{10}0+\dbinom{10}1+\cdots+\dbinom{10}9[/tex]
Without computing each term directly, let's instead use a direct result from the binomial theorem, which says
[tex]\displaystyle\sum_{n=0}^N\binom Nna^nb^{N-n}=(a+b)^N[/tex]
If we replace [tex]a=b=1[/tex], then we're left with
[tex]\displaystyle\sum_{n=0}^N\binom Nn=(1+1)^N=2^N[/tex]
We can use this to evaluate our sum directly:
[tex]\displaystyle\sum_{n=0}^9\binom{10}n=\sum_{n=0}^{10}\binom{10}n-\binom{10}{10}=2^{10}-1=1023[/tex]
