The answer is no. For $F=\{2,3\}$ (which I would rather describe as $\{0,2,3\}$) the density you want would be $\frac25$ rather than $\frac13$:
A $C$ which works for that $F$ is $$\{\,0,1,5,6,10,11\cdots\}=\{0,1\}+5\mathbb{N}$$ A little reflection may show that one can't do better. I give a proof next, though I feel there must be a better proof.
UPDATE I'm not redoing this analysis but I do sketch how one can use $14$ windows of length $4$ instead and I list them below. I'm not sure it is easier.
Your condition on $C$ is that it contain at least one member of each triple $\{x,x+2,x+3\}.$ Consider a moving window of $5$ integers $[x,x+1,x+2,x+3,x+4].$ For the set $C$ above, the intersections of $C$ with the moving window will be $0+\{0,1\}$ then $1+\{0,4\}$ then $2+\{3,4\}$ then $3+\{2,3\}$ then $4+\{1,2\}$ then $5+\{0,1\}$ etc. Ignoring $x,$ what we see in the window is a repeating sequence $\{0,1\},\{0,4\},\{3,4\},\{2,3\},\{1,2\}.$ The intersection of $C$ with each window is $\frac25$ of the window and that is the density of $C.$
There is only one possibility for the intersection of $C$ with a window having size less than two: namely $x+\{3\}.$ Any other possibility of size one misses one of $\{x,x+2,x+3\}$ and $\{x+1,x+3,x+4\}.$ To have a $C$ with density less than $\frac25$ would require seeing just $\{3\}$ in a window infinitely often and, at least sometimes, seeing it twice without, between these two occurences, ever seeing a window with three elements of $C$ in it. After one window the next one is the previous with $0$ removed, if present, every other member reduced by $1$ and perhaps $4$ added.
After $\{3\}$ we must see $\{2,4\}$ then either $\{1,3\}$ or $\{1,3,4\}.$ However $\{1,3,4\}$ is already a window with three things and after $\{1,3\}$
one must see $\{0,2,4\}.$ So the contribution to the density is not less than $\frac25$ , In fact it would be greater.
Of the $32$ possible things to see in a window only $25$ are allowed by the condition (the forbidden ones are the empty set, four singleton sets, $\{0,2\}$ and $\{1,4\}$).
UPDATE Actually it suffices to consider windows of width $4$ and have "After one window the next one is the previous with 0 removed, if present, every other member reduced by 1 and perhaps 3 added." The least dense cycle (in a sense slightly described below) involves the sequence of windows $$\{0,1\},\{0\},\{3\},\{2,3\},\{1,2\}$$ This is the same as the previous sequence except that we aren't looking at $4$. The densities are now $\frac24,\frac14,\frac14,\frac24,\frac24$ with average value (as before) of $\frac25.$
At the end I list all the possible transitions.
Here is a graph of the evolving density with two passes of your random process for $F=\{2,3\}.$ I went out to $10000$ and omit the first $2000$ or so from the graph.
I previously thought that a similar (or better) analysis should show that for $F=\{2,3,4,\cdots ,k+1\}$ the minimal density would be $\frac2{k+2}$ and this is the highest it would have to be for a set of size $k.$ But I was wrong. One can use $C=\{0,1\}+(k+3)\mathbb{Z}$ but also $C'=k\mathbb{Z}$ works. For my example of $k=2,$ $C$ is better than $C'.$ For $k=3$ it is a tie and for $k \gt 3$ $C'$ is better than $C.$ I no longer have a conjecture about how bad things can be for $|F|=k+1.$
Essentially (with some notation adjustment) the question concerns covering $\mathbb{N}$ with translates of $F,$ a set of $k+1$ integers the smallest being $0$ and the largest $m_k.$ So $$\mathbb{N} = \bigcup_{c \in C}(c+F).$$ For convenience we might allow a few misses below $2m_k$ So $$\mathbb{N} = \{0,1,\cdots,2m_k-1\} \cup \bigcup_{c \in C}(c+F).$$The question is how little overlap is possible between the translates. If , and only if, one can have a tiling with no overlaps then there is a set $C$ with density $\frac1{k+1}.$
It turns out that it is enough to consider sets $C$ of the form $C=G+ M\mathbb{N}$ where $G \subset \{0,1,\cdots,M\}$ and $M$ is an integer no larger than $2^{m_k+1}.$ So $C$ will have density $\frac{j}M$ with $j=|G|.$ Then , with no misses, $$\mathbb{Z} = F+(G+M\mathbb{Z}).$$ So density $\frac1{k+1}$ occurs precisely when $F$ tiles $\mathbb{Z}$ by translation. These sets are classified when $F$ has order $p^{\alpha} q^{\beta}$ but the general case is an intriguing problem. The first case I really don't know about is $|F|=30$
One example : $F=\{0,1,4,5,8,9\}$ with $C=\{0,2\}+12\mathbb{Z}$ of density $\frac2{12}=\frac16.$ The same $C$ works for $F'=\{0,4,8,13,17,21\}$ which is the same as $F$ $\bmod 12.$ and also for $\{0,3,7,11,16,20\}$ which is $F'$ cyclically shifted $3$ spaces to the right.
To find the minimal density and optimal choices of $G$ one can in principle make a directed graph with $2^{m_k+1}$ states corresponding to windows of width $m_k+1$ with their various densities of the form $\frac{\alpha}{m_k+1}$. An example for $\{0,2,3\}$ immediately follows.
Within this graph one wishes to find the directed cycles of minimal average density. and use one of them. A key insight is that what one does next (in building $C$ from left to right) is affected only by the most recent choices so the "best thing" to do at a certain stage is the same as it was the last time you saw the same situation. Hence the optimal path can be periodic.
$ \{0\} \rightarrow \{3\}$
$ \{2\} \rightarrow \{1, 3\}$
$ \{3\} \rightarrow \{2\},\{2,3\}$
$ \{0, 1\} \rightarrow \{0\},\{0,3\}$
$ \{0, 2\} \rightarrow \{1, 3\}$
$ \{0, 3\} \rightarrow \{2\},\{2,3\}$
$ \{1, 2\} \rightarrow \{0, 1\},\{0,1,3\}$
$ \{1, 3\} \rightarrow \{0, 2\},\{0,2,3\}$
$ \{2, 3\} \rightarrow \{1, 2\},\{1,2,3\}$
$ \{0, 1, 2\} \rightarrow \{0, 1\}, \{0, 1, 3\}$
$ \{0, 1, 3\} \rightarrow \{0, 2\},\{0,2,3\}$
$ \{0, 2, 3\} \rightarrow \{1, 2\}, \{1, 2, 3\}$
$ \{1, 2, 3\} \rightarrow \{0, 1, 2\}, \{0, 1, 2, 3\}$
$ \{0,1, 2, 3\} \rightarrow \{0, 1, 2\}, \{0, 1, 2,3\}$
