As noted in Dylan's comment, such a subgroup may not exist. The smallest example is the non-existence of a 6-element subgroup of $A_4$.
There is a systematic method for deciding whether there is such a subgroup, and constructing one if one exists, but it's rather trivial. First, see if there is an element of order $k$. If so, you know what to do. If not, pick two elements of order dividing $k$, and see whether they generate a subgroup of order $k$. If they do, you win. If not, pick a different pair of elements of order dividing $k$. Continue, systematically, if necessary until you have tested all pairs of elements (your group is finite, so this terminates). Then, if necessary, start on all sets of three elements, each of order dividing $k$, then all sets of four elements, etc. By this exhaustive and systematic procedure you will either find a subgroup or prove that there isn't one.
I make no claim for efficiency. In practice, you will find shortcuts. E.g., if $\lbrace a,b\rbrace$ generates a group of order not dividing $k$, then you don't have to test $\lbrace a,b,c\rbrace$. Also, you never have to test a set of more than $k$ elements.
Come to think of it, you could just go through the subsets of size $k$, testing each in turn to see whether it's a subgroup, stopping either when you find one that is or when you've eliminated all of them. Highly inefficient, but unarguably systematic.