The following is a construction of an analytic set $A\subseteq\omega^\omega\times\omega^\omega$ whose vertical sections exclude at most a single point, and which does not admit a Borel uniformization. This is as stated in jonathanverner's answer, although that did not include a construction - only mentioning exercise 36.25 in Kechris: *Classical Descriptive Set Theory*. The construction below is after viewing the hint in Kechris's book. I'll use the notation $\mathcal{N}$ for Baire space $\omega^\omega$.

**Construction**. Let $S$ be a universal subset of $\mathcal{N}\times\mathcal{N}^3$ (i.e., $S$ is closed and every closed subset of $\mathcal{N}^3$ is a vertical section $S_x$ of $S$ for some $x\in\mathcal{N}$). Then, let $A\subseteq\mathcal{N}^2$ consist of the points $(x,y)$ such that whenever, for any such $x$, there is a *unique* $(u,v)\in\mathcal{N}^2$ with $(x,x,u,v)\in S$ we have $y\not=u$. The set $A$ satisfies the required properties.

It is clear the sections $A_x$ exclude at most a single point. More precisely, it excludes the point $y$ if and only if there is a unique $(u,v)\in\mathcal{N}^2$ with $(x,x,u,v)\in S$ and $y=u$.

**$A$ does not have a Borel section:** Suppose that $\Gamma\subseteq\mathcal{N}^2$ is Borel. I'll now use the fact that every Borel subset of a Polish space is a continuous bijective image of a closed subset of $\mathcal{N}$ (Kechris, Theorem 13.7). By taking the graph of such a continuous bijection, there exists a closed set $C\subseteq\mathcal{N}^2\times\mathcal{N}$ whose projection onto $\mathcal{N}^2$ is one-to-one and has image $\Gamma$. Write $C=S_x$ for some $x\in\mathcal{N}$. Then, $(x,y)\in\Gamma$ if and only if $(x,x,y,v)\in S$ for some $v$, which is then unique. If $\Gamma_x$ consists of the single point $y$ then $(y,v)$ is unique such that $(x,x,y,v)\in S$ and, by construction, $(x,y)\not\in A$. So, $\Gamma$ is not a section of $A$.

**$A$ is analytic:** The set $T=\lbrace(x,y,z)\colon(x,x,y,z)\in S\rbrace$ is a closed subset of $\mathcal{N}^3$ and $(x,y)\in A$ iff whenever there is unique $(u,v)\in\mathcal{N}^2$ with $(x,u,v)\in T$ we have $y\not=u$. We can write $A=B\cup(\mathcal{N}^2\setminus C)$ where
$$
\begin{align}
&B = \left\lbrace(x,y)\in\mathcal{N}^2\colon(x,y^\prime,z)\in T, {\rm some\ }y^\prime\not=y\right\rbrace\\
&C=\left\lbrace(x,y)\in\mathcal{N}^2\colon\exists ! z\in\mathcal{N}{\rm\ with\ }(x,y,z)\in T\right\rbrace.
\end{align}
$$
As $B$ is the projection onto $\mathcal{N}^2$ of the (Borel) set of $(x,y,y^\prime,z)\in\mathcal{N}^2\times\mathcal{N}^2$ with $(x,y^\prime,z)\in T$ and $y\not=y^\prime$, it is analytic. Finally, $\mathcal{N}^2\setminus C$ is analytic by the following (surprising) result from Kechris (originally by Lusin).

**Theorem** (Kechris, Thm 18.11). Let $X,Y$ be standard Borel spaces and $B\subseteq X\times Y$ be Borel. Then,
$$
\left\lbrace x\in X\colon\exists ! y\in Y{\rm\ s.t.\ }(x,y)\in B\right\rbrace
$$
is coanalytic.