Game schedule where each player meets only once

n = 7;
k = 3;

DeleteDuplicates[Subsets[Range @ n, {k}], Length[Intersection[##]] >= 2 &]

{{1, 2, 3}, {1, 4, 5}, {1, 6, 7}, {2, 4, 6}, {2, 5, 7}, {3, 4, 7}, {3, 5, 6}}

Update: Although DeleteDuplicates approach replicates the result of For loops in OP, both are greedy approaches and we don't have any guarantee that we get a schedule of maximum possible length. To get a schedule of maximum possible length we can use Roman's approach. The following is an alternative specification of the input graph that is slightly faster:

grph = RelationGraph[Length[Intersection[##]] <= 1 &, Subsets[Range @ n, {k}]];

allscheadules = FindClique[grph , ∞, All];

Length @ allscheadules

450

Max[Length /@ allscheadules]

7

So, for {n, k} = {7, 3} the schedule we get using the greedy approach cannot be made any longer.

For {n, k} = {9, 3} the two approaches give substantially different results:

{n, k} = {9, 3};

s1  = DeleteDuplicates[Subsets[Range @ n, {k}], Length[Intersection[##]] >= 2 &]

{{1, 2, 3}, {1, 4, 5}, {1, 6, 7}, {1, 8, 9}, {2, 4, 6}, {2, 5, 7}, {3, 4, 7}, {3, 5, 6}}

Length @ s1

8

Tally @ Sort @ Flatten @ s1

{{1, 4}, {2, 3}, {3, 3}, {4, 3}, {5, 3}, {6, 3}, {7, 3}, {8, 1}, {9, 1}}

grph = RelationGraph[Length[Intersection[##]] <= 1 &, Subsets[Range @ n, {k}]];

allscheadules = FindClique[grph , ∞, All];

Length @ allscheadules

834960

Max[Length /@ allscheadules]

12

s2 = MaximalBy[allscheadules, Length, 1]

{{{1, 2, 9}, {1, 3, 8}, {1, 4, 7}, {1, 5, 6}, {2, 3, 7}, {2, 4, 6}, {2, 5, 8}, {3, 4, 5}, {3, 6, 9}, {4, 8, 9}, {5, 7, 9}, {6, 7, 8}}}

Tally @ Sort @ Flatten @ s2

{{1, 4}, {2, 4}, {3, 4}, {4, 4}, {5, 4}, {6, 4}, {7, 4}, {8, 4}, {9, 4}}

n = 7;
k = 3;

list all possible games:

g = Subsets[Range[n], {k}];

Find a maximal-size clique of games that don't overlap:

First@FindClique[AdjacencyGraph[Outer[Boole[Length[Intersection[##]] <= 1] &, g, g, 1]]]
(*    {1, 10, 15, 21, 24, 28, 29}    *)

Which games are these?

g[[%]]
(*    {{1, 2, 3}, {1, 4, 5}, {1, 6, 7}, {2, 4, 6}, {2, 5, 7}, {3, 4, 7}, {3, 5, 6}}    *)

This method gives the same result as @kglr's but is much slower, so I don't recommend using it. You can view it as a proof of the other method.