Is the chance of breaking even in this coin toss game $43.75\%$?

No matter what your betting strategy is, your expected number of coins will always remain 1000 coins. This includes the faint possibility that you cannot even make your 100 games because you lost your 1000 coins already after 28 rounds.

Given this fact, what is the probability to have more or less than 1000 coins after very many (say, $N$) games? If $p_n$ denotes the probability to have $n$ coins after $N$ games, then the expected value is $$E(X)=1000=\sum_{n=0}^\infty n p_n$$ And the probability of having less or more than 1000 coins is $$p_{<1000}=\sum_{n=0}^{999} p_n$$ $$p_{>1000}=\sum_{n=1001}^{\infty} p_n.$$ With just a few games, we have a high probability of having slightly more than 1000 coins and a low probability of having much less than 1000 coins.

However, for large $N$, when it is possible to have gained much, much more than the original 1000 coins ans also not too unlikely that we may have gone bancrupt with $0$ coins (and canot afford to bet more than $0$ coins) the situation changes: $p_{<1000}$ is largely dominated by $p_0$, which makes no contribution at all to $E(X)$, whereas $p_n$ for large $n$ are non-zero. This is like the oppsoite situation of what the betting strategy intended: There is a high probability of a moderate loss (of 1000 coins) and a small probability of substancial gain. Especially, for sufficiently large $N$, we will have $p_{<1000}>p_{>1000}$. It would be interesting to know, at which number $N$ of rounds the switch occurs. I estimate that it will take more than your suggested 100 games, though.