It is well-known that the space of holomorphic functions $\mathcal{O}(U)$ (with the standard topology of compact-uniform convergence) on an open set $U \subset \mathbb{C}$ is a projective limit of Banach spaces $\mathcal{A}(K)$, where $\mathcal{A}(K)$ consists of functions holomorphic on the interior and continuous on the boundary of $K$.
I want to represent $\mathcal{O}(U)$ as a projctive limit of Hilbert spaces. I thought for a second that I've solved it but then it turned out I wasn't even close.
My idea was to use Hardy spaces (because those are the only Hilbert spaces related to holomorphic functions I can think of) to represent $\mathcal{A}(K)$ as a projective limit and then $\mathcal{O}(U)$ as the limit of the limits. For this I wanted to cover the interior of $K$ by open balls and then consider Hilbert spaces of functions on them and Hilbert direct sums for the unions of balls. Then I realised that that Hardy space is defined only for the unit ball centred at origin so I can't use them for my idea.
Maybe I can take $L^2$-norm on $\mathcal{A}(K)$ and take completion by I doubt it'll work.
Reduction to writing $\mathcal O(U)$ as a projective limit of prehilbert spaces
Notice that since $\mathcal O(U)$ is Hausdorff and complete, if you are able to write it as a projective limit of prehilbert spaces, then you are able to write it as a projective limit of Hilbert spaces by taking the projective limit of the Hausdorff completions of the prehilbert spaces.
This is where the "not very concrete" aspect of my comment comes from: I will show that there is a projective system of prehilbert spaces of which $\mathcal O(U)$ is the projective limit, but I will not try to give a nice description of the Hausdorff completions of the prehilbert spaces (I do not know if there is a satisfying one).
Reduction to finding prehilbert norms defining the topology of $\mathcal O(U)$
An easy way of showing a locally convex space is a projective limit of prehilbert spaces is to show that its topology can be defined by a family of prehilbert norms (norms that are induced by inner products).
Indeed, let $E$ be a locally convex space whose topology is defined by a family $(p_\alpha)_{\alpha\in A}$ of prehilbert norms. For every finite subsets $B$ and $C$ of $A$ such that $C$ contains $B$, denote by $E_B$ the normed space whose underlying vector space is the same as $E$ and whose norm is $$p_B:x\mapsto\biggl(\sum_{\alpha\in B}(p_\alpha(x))^2\biggr)^{1/2}$$ and by $i_{BC}:E_C\to E_B$ the identity mapping. The spaces $E_B$ are prehilbert spaces, the mappings $i_{BC}$ are linear and continuous, and together, they form a projective system of which $E$ can be checked to be the projective limit.
So showing that $\mathcal O(U)$ is a projective of Hilbert spaces boilds down to showing its topology can be defined by a family of prehilbert norms.
Description of prehilbert norms defining the topology of $\mathcal O(U)$
The most obvious way to define the topology of $\mathcal O(U)$ by a family of seminorms is to use the family of the norms of the form $$ q_K:f\mapsto\sup_{z\in K}|f(z)| $$ as $K$ ranges overs the set of all compact subsets of $U$. But it is also possible to define the topology of $\mathcal O(U)$ using only the norms of the form $q_D$ as $D$ ranges overs the set of all compact disks of the complex plane with positive radius and contained in $U$, because any compact subset $K$ of $U$ can be covered by finitely many such disks $D_1,D_2,\dots,D_n$, which then gives $q_K\leq\sup(q_{D_1},q_{D_2},\dots,q_{D_n})$.
For every compact disk $D$ with positive radius and contained in $U$ and for every $f\in\mathcal O(U)$, let $p_D(f)$ be the $\mathrm L^2$ norm of $f$ with respect to the usual complex Radon measure on $\partial D$ associated to the functional $$ \begin{array}{ccl}\mathcal C(\partial D)&\longrightarrow&\mathbf C\\ \varphi&\longmapsto&\displaystyle\int_{\partial D}\varphi(\lambda)\,d\lambda\end{array}. $$ The functions $p_D$ are prehilbert norms $\mathcal O(U)$. I will show that they define the topology of $\mathcal O(U)$ by comparing them with the norms $q_D$, which will conclude my answer.
First, it is quite clear that if $D$ is a compact disk contained in $U$ with radius $r>0$, the inequality $p_D\leq 2\pi r\mathop{q_D}$ always holds. Conversely, let $D_0$ be a compact disk contained $U$ with radius $r_0>0$. Let $D_1$ be another compact disk contained in $U$, with the same center as $D_0$ and radius $r_1>r_0$. The Cauchy integral formula and the Cauchy-Schwarz inequality show that for every $f\in\mathcal O(U)$, we have \begin{align*} q_{D_0}(f)=\sup_{z\in D_0}|f(z)|&=\sup_{z\in D_0}\Biggl|\frac 1{2\pi i}\int_{\partial D_1}\frac{f(\lambda)}{\lambda-z}\,d\lambda\Biggr|\\ &\leq\sup_{z\in D_0}\frac 1{2\pi}\Bigl\|\frac 1{\cdot-z}\Bigr\|_{\mathrm L^2(\partial D_1)}\bigl\|f\bigr\|_{\mathrm L^2(\partial D_1)}\leq\frac 1{2\pi}\frac{(2\pi r_1)^{1/2}}{r_1-r_0}\mathop{p_{D_1}}(f). \end{align*}