Karush-Kuhn-Tucker in infinite dimension
Yes, Bachir et al. (2021) extend the Karush-Kuhn-Tucker theorem under mild hypotheses, for an infinite number of variables (their Corollary 4.1).
I give hereafter a weaker version of the generalization of Karush-Kuh-Tucker in infinite dimension:

Let $X\subset\mathbb{R}^{\mathbb{N}}$ be a nonempty convex subset of
$\mathbb{R}^{\mathbb{N}}$ and let $x^{*}\in Int\left(X\right)$. Let
$f,g_{1},g_{2},...,g_{m}:X\rightarrow\mathbb{R}$ be convex functions
continuous at $x^{*}$ and term-to-term differentiable at $x^{*}$, i.e.
such that the functions
$f_{n,x^{*}}\left(x_{n}\right):=f\left((x_1^{*},...,x_{n-1}^*,x_n,x_{n+1}^*,...)\right)$ and
$g_{j,n,x^{*}}\left(x_{n}\right):=g_{j}\left((x_1^{*},...,x_{n-1}^*,x_n,x_{n+1}^*,...)\right)$ are
differentiable at $x_{n}$ for all $n\in\mathbb{N}$ and
$j\in\left\{ 1,2,...,m\right\}$.
(Qualification condition) Suppose that for all $k\in\mathbb{N}^{*}$
and for all $x\in X$,
$$x^{*}+P^{k}\left(x-x^{*}\right)=\left(x_{1},...,x_{k},x_{k+1}^{*},x_{k+2}^{*},...\right)\in X$$
If there exist
$\left(\lambda_{j}^{*}\right)_{j}\in\left(\mathbb{R}_{+}\right)^{\mathbb{N}}$
such that
$$\lambda_{j}^{*}g_{j}\left(x^{*}\right)  =0,\:\forall j\in\left\{
1,2,...,m\right\} \quad \quad \quad \quad \quad (1)$$
$$f_{n,x^{*}}^{\prime}\left(x_{n}^{*}\right)+\sum_{j=1}^{m}   \lambda_{j}^{*}g_{j,n,x^{*}}^{\prime}\left(x_{n}^{*}\right)=0,\:\forall
n\in\mathbb{N} \quad \quad (2)$$
(Sufficiency) Then $x^{*}$ is an optimal solution on $\Gamma:=\left\{
\left(x_{i}\right)_{i}\in
X\,:\,g_{1}\left(x\right)\leq0,...,g_{m}\left(x\right)\leq0\right\} :$
$$f\left(x^{*}\right)=\underset{x\in\Gamma}{\inf}f\left(x\right)$$
(Necessity) Besides, if $x^{*}$ is an optimal solution on $\Gamma$ and
if the Slater condition $Int\left(\Gamma\right)\neq\emptyset$ is
verified, then there exist unique
$\left(\lambda_{j}^{*}\right)_{j}\in\left(\mathbb{R}_{+}\right)^{\mathbb{N}}$
which verify the (Karush-Kuhn-Tucker) conditions (1) and (2).

The number of constraints has to be finite, but simple constraints like non-negativity constraints can be replaced by an equivalent restriction on the domain of the variables. For example, instead of the constraints $\forall n \in \mathbb{N},\;x_n \geq 0$ on the domain $\mathbb{R}^{\mathbb{N}}$, one can take $X=(\mathbb{R}_+)^{\mathbb{N}}$, and the theorem applies.
Note that the (sufficiency) result is easy to prove when one further assumes that the convex Lagrangian $\mathcal L(x, \lambda)=f(x)+\sum_{j=1}^m\lambda_j g_j(x)$ is Gateaux differentiable, with a Gateaux derivative equal to 0 at $u=(x^*, \lambda^*)$.
Indeed, a function $h: V \rightarrow \mathbb{R} $ convex and Gateaux differentiable on $V$ verifies $h(v)-h(u) \geq h^\prime(u; v-u), \forall u,v \in V$, where $h^\prime(u; v)$ is the directional derivative of $h$ at $u$ in the direction $v$. (One can see that from the definition of convexity: $h(u)+\theta \left( h(v) -h(u) \right) \geq h\left(u+\theta (v-u)\right)$; subtracting $h(u)$, dividing by $\theta$, and taking the limit when $\theta \rightarrow 0^+$; see this for more details). Applying that inequality to the Lagrangian at $u$ proves that the Lagrangian admits a minimum at $u$, which solves the minimization program: $ f(x^*) =L(x^*, \lambda^*) \leq f(x)+\sum_{j=1}^m\lambda_j g_j(x) \leq f(x), \; \forall x\in \Gamma$.
However, in general, it is not easy to prove that the Gateaux derivative of a convex series (such as an infinite Lagragian) (exists and) equals 0 at some point $u$, unless one uses the result (Theorem 3.14 in Bachir et al. (2021)) that the Gateaux derivative is thus equal to the sum of derivatives of each term in the series.