I'm a bit confused on the choice of sets that authors choose and why. For example :
"Any harmonic function $u$ on an open subset $\Omega$ of $R^2$ is locally the real part of a holomorphic function."
Also
"Given $h$ positive, $(a,b) \in R^2$. Let $U$ be harmonic on $I=(a-h,a+h) \times (b-h, b+h)$. Then there exists $f$ analytic on $\sigma(I)$ such that $U=Re\ F$, $F$ being the associate of $f$"
I'm not sure what restrictions are necessary? It seems like it would be true for any open set of any size (connected or otherwise). Since the harmonic and analytic definitions are local in nature, I'm not sure why it is called a "local" converse, or what that really means in this context? Also, I'm not sure why the second author chose the set he did? I feel like I am missing something.
I just post my comment as an answer because it has grown too long.
If you are given any holomorphic function, its real and imaginary part are harmonic (rewrite the Laplacian in complex coordinates using Wirtinger's equations).
Conversely, if you are given any harmonic function and you want to construct a holomorphic function with the given harmonic function as real or imaginary part, there is an integral involved and this imposes the condition of simple connectedness on the defining domain of the harmonic function.
I'm not entirely sure wether this is the end of the story. The connection between holomorphic and harmonic functions plays an important role in what is called Levi's problem. You find a complete and nice overview of this topic in Grauert-Fritzsche: From holomorphic functions to complex manifolds, Chapter II. This is almost 60 pages because the topic is, as many things in complex analysis, in parts very technical.
EDIT: According to your comment, here are some more details on where we need a simply connected domain.
Let $G\subset \mathbb C$ be any domain. By a domain I mean a connected, non-empty open subset.
Definition. Let $u,v:G\to \mathbb R$ be harmonic functions. If the function $f=u+iv$ is holomorphic on $G$ we call $(u,v)$ an adjoint harmonic pair.
We need a simple connected domain in the proof of the following theorem.
Theorem. If $G$ is simply connected and $u:G\to \mathbb R$ harmonic, then there is a harmonic function $v:G\to \mathbb R$ such that $(u,v)$ is an adjoint harmonic pair. In other words, every harmonic function on a simply connected domain is the real part of some holomorphic function.
Proof. Notice that in complex coordinates, $$\Delta=4\partial \overline \partial = 4\overline \partial \partial.$$ Set $g:=\partial u$. It follows from $\Delta=4\partial \overline \partial = 4\overline \partial \partial$ that $g$ is holomorphic on $G$. Since $G$ is simply connected there exists a primitive function $f$ of $g$. Using the Cauchy Riemann equations and Wirtinger calculus we see that \begin{align} g=\partial f&=\frac{1}{2}\left(\partial_x-i\partial_y\right)(\Re(f)+i\Im(f))\\ &=\frac{1}{2}(\partial_x \Re(f)-i\partial_y\Re(f)+i(\partial_x\Im(f)-i\partial_y\Im(f)))\\ &=\frac{1}{2}(\partial_x \Re(f)-i\partial_y\Re(f)-i\partial_y\Re(f)+\partial_x\Re(f)))\\ &=\partial \Re(f), \end{align} where $\Re,\Im$ denote the real, resp. imaginary part. Moreover \begin{equation*} \partial (u-\Re(f))=g-g=0, \end{equation*} hence $u-\Re(f)=c\in \mathbb C$. As $f$ is only unique up to a constant we can, if necessary, change $f$ to $f-c$ and the claim follows.