One can show that $\Pi^1_1$-determinacy is equivalent to the existence of $0^\sharp$ (left to right is due to Harrington, and right to left is due to Martin)., but every proof of "$\Pi^1_1$-determinacy implies that $0^\sharp$ exists" that I know of goes through the following lemma:
$\mathbf{Lemma}$: If there is a real $x$ such that every $x$-admissible ordinal is a cardinal in $L$, then $0^\sharp$ exists.
Edit:
I've realized that this is actually a rather difficult question to answer as stated. I asked it because I could not have been the first person to think of this, and was curious if anyone else had made progress. However, the question was ill-formed in some sense, but here is related question that is perhaps more answerable:
As far as I know, there is no proof of "$0^\sharp$ exists from $\Pi^1_1$-determinacy that does not utilize the above lemma. Is this simply because no one has worked on this since, or because there is something about $\Pi^1_1$ sets that make this difficult?
From what I understand, $0^\sharp$ is a well-studied object insofar as much is known about it. There is also a fair bit of work that has been done on determinacy, but I have not seen as much in conjunction with $\Pi^1_1$ determinacy. Most of what I have seen is in regards to OD-Determinacy, Projective Determinacy, and Determinacy in $L(\mathbb{R})$. Now, if one were interested in trying to produce a model that contains $0^\sharp$ from a $\Pi^1_1$ game, the next question is whether or not the above technique of Martin and Solovay would be the best approach. This technique was used with stronger determinacy hypotheses, so I wonder if the low complexity of $\Pi^1_1$ sets would make it the case that a different approach would be better.
A good starting place may be the known proofs of "$0^\sharp$ exists from $\Pi^1_1$-determinacy. We define a set $A\subset{}^\omega\omega$ as follows: $a\in A$ if, and only if there is a binary relation $R$ on $\omega$ which is recursive in $a$ such that $(\omega, R)$ is isomorphic to an end extension of $(L_{\omega_1(a)},\in)$ where $\omega_1(a)$ denotes the first $a$-admissible ordinal. We then prove that $A$ is $\Sigma^1_1$, closed under Turing equivalence, and non-empty. This is the game that Harrington used to prove that $0^\sharp$ exists. However in an exercise in Martin's notes on determinacy, he defines a different game in the hint.
We play a game $G$ on $^{<\omega}\omega$. For each play of $G$, $I$'s part codes a relation $R$ on $\omega$ and let $II$'s part of the play code a relation $E$ on $\omega$. If $R$ is not a well-ordering of $\omega$, $I$ loses. If $R$ is a wellordering of $\omega$, let $\beta$ be its order type. Then, $II$ wins if and only if $(\omega;E)$ is a model of extensionality, and there is a $g:L_\beta\rightarrow\omega$ that embeds $(L_\beta;\in)$ into $(\omega;E)$ as an initial segment. We then show that the set we are playing over is $\Pi^1_1$, and that $I$ has no winning strategy in $G$.
This seems a bit closer to the idea presented in the first real paragraph of this question. In particular, perhaps one could construct a $\Pi^1_1$ game that codes a nontrivial elementary embedding $j:L_\alpha\prec L_\beta$ for limit ordinals $\alpha$ and $\beta$ using a similar idea.