Can two queens win against a knight on an infinite chess-board?

Inspired by this question :

A simple game on infinite chessboard

I ask whether the special case ($0$ bishops) will already be sufficient.

More concretely : The game is as follows. Player $A$ places two queens, then player $B$ places a knight on an infinite chess-board. Then, the queens and the knight move alternately. The knight is allowed to capture a queen. If the knight can be checkmated (that is it is attacked and has no way out), then player $A$ wins. If the knight survives forever, player $B$ wins. If the knight is not attacked , but cannot move, we have a stalemate, which is a draw.

I think, the queens can win against best defence. First step is to bring the queens near enough to the knight, second step is to attack the knight from two opposite directions. After trying some constellations, I am convinced that the queens will win assuming best play of both sides.

How can we prove this ?

I found a surefire way to win the game in at most 8 moves (not including placing the queens) , even if we lose the game in the case the knight can capture a queen.

$(1)$ Place one queen anywhere and the other directly right from it. This prevents the knight from forking the queens.

$(2)$ Move one queen far left from the knight (lets say $100$ lines) and the other far right from the knight (lets say $100$ lines). Begin with the attacked queen, if the knight attacks a queen after being placed.

$(3)$ Attack the knight from left or right with one of the queens.

a) The knight moves 2 lines WLOG above this line. Then place the other queen 2 lines higher.

b) The knight moves 1 line WLOG above this line. Then attack the knight from left/right with the other queen. Whatever the knight does. we can achieve the constallation that one queen is 2 lines above and the other 2 lines below the knight. This phase requires at most $3$ moves.

$(4)$ The knight must now move to a line which is 1 line below or above one of the queens. We attack the knight with this queen directly from above or below.

$(5)$ We can always achieve this configuration (upto symmetries)

$$ Q\ X\ X $$ $$ X\ N\ X $$ $$ X\ X\ X $$ $$ X\ X\ X $$ $$ X\ X\ Q $$

$(6)$ It is easy to see that the knight will be checkmated in the next move.

In total, we only need 8 moves and this can probably be improved.