See if you can find the paper, I N Schnurnikov, Into how many regions do $n$ lines divide the plane if at most $n-k$ of them are recurrent? The abstract says, inter alia,
Thus, a new proof of N. Martinov’s theorem is obtained. This theorem determines all pairs of integers $(n, f)$ such that there is an arrangement of $n$ lines dividing the projective plane into $f$ regions.
Or, if you read Bulgarian, look for Nikola Martinov, Solution of a Federov-Grünbaum problem, Annuaire Univ. Sofia Fac. Math. Inform. 87 (1993), no. 1-2, 73–85 (1999), MR1745340 (2000k:52020).
Wait a minute - it's also in English: Nicola Martinov, Classification of arrangements by the number of their cells, Discrete Comput. Geom. 9 (1993), no. 1, 39–46, MR1184692 (93g:52008).
EDIT 17 May: See also Oleg A Ivanov, On the number of regions into which $n$ straight lines divide the plane, Amer Math Monthly 117 (Dec. 2010) 881-888. Curiously, this paper does not reference Martinov's work.
For the question stated in the title, the answer is yes, if more is interpreted as "more than or equal to".
Proof: let $\Lambda$ be a collection of lines, and let $P$ be the extended two plane (the Riemann sphere). Let $P_1$ be a connected component of $P\setminus \Lambda$. Let $C$ be a small circle entirely contained in $P_1$. Let $\Phi$ be the conformal inversion of $P$ about $C$. Then by elementary properties of conformal inversion, $\Phi(\Lambda)$ is now a collection of circles in $P$. The number of connected components of $P\setminus \Phi(\Lambda)$ is the same as the number of connected components of $P\setminus \Lambda$ since $\Phi$ is continuous. So this shows that for any collection of lines, one can find a collection of circles that divides the plane into at least the same number of regions.
Remark: via the conformal inversion, all the circles in $\Phi(\Lambda)$ thus constructed pass through the center of the circle $C$. One can imagine that by perturbing one of the circles somewhat to reduce concurrency, one can increase the number of regions.
Another way to think about it is that lines can be approximated by really, really large circles. So starting with a configuration of lines, you can replace the lines with really really large circles. Then in the finite region "close" to where all the intersections are, the number of regions formed is already the same as that coming from lines. But when the circles "curve back", additional intersections can happen and that can only introduce "new" regions.
Lastly, yes, the number you derived is correct. See also this OEIS entry.
Best Answer
In order to maximize the number of regions, it seems you want to make sure that
Depending on how rigorous your proof has to be, you might have to elaborate on this point.
So you could choose $n$ points on the interior of the crescent, and draw a tangent at each. Then you consider the whole thing as a graph in terms of its Euler characteristic formula:
Every straight line intersects all other lines and also touches the inner boundary. So every one of the $n$ lines contributes $n+1$ straight line segments as edges of the graph. Furthermore, the inner arc is intersected once by every line, so it is split into $n+1$ curved edges. The outer arc is intersected twice, resulting in $2n+1$ curved edges. So the total number of edges is
$$E=n(n+1)+(n+1)+(2n+1)=n^2+4n+2$$
Every point of intersection between two lines is a vertex, amounting to $\binom{n}{2}=\frac{n(n-1)}2$ vertices inside the crescent. Then there are $n$ points where the lines intersect the inner arc, and $2n$ where they intersect the outer. The tips where inner and outer arc meet contribute two more vertices. So the total number is
$$V=\binom n2+n+2n+2=\frac{n^2+5n+4}{2}=\frac{(n+1)(n+4)}{2}$$
From these two you can deduce the number of faces, not counting the one outside the crescent, as
$$F=\chi+E-V=1+E-V=\frac{n^2+3n+2}{2}=\frac{(n+1)(n+2)}{2}$$
This is because topologically your crescent is a disk, so it has Euler characteristic $\chi=1$.
Except for an offset of one in the indices (i.e. the value of $n$), these are exactly the triangular numbers. They do confirm the numbers you gave in your question, so you did count right.