Lead against a confident slam

This is a hand from the PanIIT inter IIT bridge tournament conducted online in 2020/2021. 

Both vul, IMPS. You hold 8x, A9x, Q7xxxx, K9.

Partner passes, RHO opens 1S, you pass. LHO bids 4NT, RHO shows two keycards without Q and LHO bids 6S.

What would you lead?

Given that you hold HA, DQ and CK and LHO jumped directly to 4NT, it is very likely they have a distributional hand with a long and good club suit. The club finesse declarer needs is working, even if they can only take it once.

At the table I tried the lead of C9 from K9. This turned out to be a lucky lead as you can see from the four hands below.

IMPS  Both	North ♠ AQ7 ♥ J ♦ K ♣ AQJ76532
West ♠ 84 ♥ A92 ♦ Q76532 ♣ K9		East ♠ T62 ♥ K87643 ♦ T98 ♣ 4
	South ♠ KJ953 ♥ QT5 ♦ AJ4 ♣ T8

W	N	E	S
		P	1S
P	4NT	P	5H
P	6S	P	P
P

Declarer decide to spurn the club finesse and went up with the CA and could no longer make the contract!

Cover the plane with strips

 In the 2D plane, a strip is the space between (and including) two parallel lines (which extend to infinity in both directions). The width of a strip is the distance between the two parallel lines that make up a strip. The width of a strip is non-zero (otherwise we just call it a line).

Can you fully cover the 2D plane with strips the sums of whose widths is finite? (Proof required of course).

Scroll down for a solution.

Consider a circle of radius $R$. A strip of width $w_i$ can cover an area at most $2w_iR$.

So strips of total width $w$ (infinite number of strips or not), can cover at most an area of $2wR$.

For sufficiently large $R$, $\pi R^2 \gt 2wR$. So we cannot cover the plane with strips the sum of widths of which is finite.