Nice! The above proof proves that any configuration of 10 dots can be covered. What you have here is an example of the probabilistic method, which uses probability but gives a certain (not a probabilistic) conclusion (an example of probabilistic proofs of non-probabilistic theorems). This proof also implicitly uses the linearity of expectation, a fact that seem counter-intuitive in some cases until you get used to it.
To clarify the proof: given any configuration of 10 dots, fix the configuration, and consider placing honeycomb-pattern disks randomly. Now, what is the expected number $X$ of dots covered? Let $X_i$ be 1 if dot $i$ is covered, and $0$ otherwise. We know that $E[X] = E[X_1] + \dots + E[X_{10}]$, and also that $E[X_i] = \Pr(X_i = 1) \approx 0.9069$ as explained above, for all $i$. So $E[X] = 9.069$. (Note that we have obtained this result using linearity of expectation, even though it would be hard to argue about the events of covering the dots being independent.)
Now, since the average over placements of the disks (for the fixed configuration of points!) is 9.069, not all placements can cover ≤9 dots — at least one placement must cover all 10 dots.
Here's an $16$-approximation, assuming the object does not have holes. In particular, this approximation gives disjoint rectangles, and thus, is an instance of a Polygon Covering (nevertheless the bounds hold).
The union of the squares can be alternatively seen as a rectilinear polygon. We let the vertices of the polygon be the sharp points on the boundary of the polygon. Let $V$ be the number of such vertices. I will describe an approach based on the sweep-line triangulation algorithm. Basically, I will give a trivial triangulation if the polygon is monotone, and will later describe how to transform the polygon into monotone pieces.
If the polygon is monotone and does not have holes
First, let's consider what happens if the polygon is monotone (in $y$ coordinates) and have no holes.
Let $P$ be the optimal number of rectangles to cover this polygon. First, I claim that
$$4P \le V$$
That is, $V$ is at most $4$ times the number of rectangles in the optimal answer.
To see this, consider any rectangle covering the object. What is the maximum number of vertices this rectangle can touch? The answer is $4$ and thus, the bound follows.
Now, we can construct $V$ rectangles that cover everything as follows: Let $Y$ be the sorted list of distinct $y$ coordinates of the vertices of the polygon. For every pair of adjacent $y_i \in Y, y_{i+1} \in Y$, construct a rectangle to cover the polygon. Clearly this constructs at most $V$ rectangles.
Making the polygon monotone
We do a similar approach with triangulation: we make the polygon monotone and cover each monotone piece by rectangles. We can do monotonizing in a similar approach as in triangulation (alluded in the wiki): whenever we encounter a concave point that "faces down", we draw a line from the point left and right until we hit the boundaries of the polygon, and cut the polygon there, creating separate polygons in the process (it really is similar to the algorithm described in the wiki, but instead of connecting it to the vertex in the left/right sides of the polygon, we create a new vertex in the corresponding boundary). Thus, each "concave" point may create at most $2$ new vertices, and the $12$-approximation follows.
Now, how many concave points can a rectilinear polygon has? It is at most the number of convex points it have (in fact, it is exactly $4$ more than that). Now, if $V_x$ is the number of convex vertices of the polygon, using a similar reasoning as before, we have $4P \le V_x$, and thus, this is a $4 \times (1+3) = 16$-approx algorithm.
What if the polygon has holes?
The bounds on the optimal answer no longer holds, and thus, no guarantee is present even if we were to monotonize the polygon using the vertices of the holes.
Best Answer
I remember that problem from university! (But I didn't solve it, somebody gave me a big clue)
The answer is that it is impossible, you can't cover the circle of diameter 20 using only 19 strips $1\times 20$.
To see why, suppose you have such cover and imagine a sphere of diameter 20 cut in half by our circle. Find the orthogonal projection of each strip on the sphere, it is a ring, and it is easy to compute it's area $2\pi R x$, where $R$ is the radius of the sphere and $x$ the width of the strip.
But between all the rings, they cover the whole sphere so the total area must be at least $4\pi R^2$, so $$ N \times 2 \pi R x \ge 4 \pi R^2 $$ where $N$ is the number of strips. Letting $N=19, R=10$ and $x=1$ we obtain a contradiction.