It is well-known that if a graph has maximum degree $d$, then it is $d+1$ colorable. Say we have $d+1$ graphs $G\_1,\ldots, G\_{d+1}$ **on the same vertex set** $V$, and say each $G\_i$ has maximum degree at most $d$.


A *coloring* of $\textbf{G}:=\{G\_1,\ldots, G\_{d+1}\}$ is just a labelling of the common vertex set of $\textbf{G}$ with $\{1,\ldots, k\}$, for some $k\in\mathbb{N}$. This coloring is *proper* if for any $i\in\{1,\ldots, k\}$, no edge of $G\_i$ can be found between two vertices colored $i$.


My question is whether $\textbf{G}$ admits a proper coloring with just $d+1$ labels.


Another way to formulate this is that we are looking for a partition of the common vertex set $V=V\_1\cup \ldots \cup V\_{d+1}$, where $V\_i$ is an independent set in $G\_i$.


The simplest case is when $d=1$. In this case, $\textbf{G}=\{G\_1, G\_2\}$, and both $G\_1$ and $G\_2$ are matchings. I claim that $\textbf{G}$ can be properly $2$-colored. Proceed with induction - base case is clear. If there exists a vertex that is isolated in $G\_1$, we can remove that vertex, apply induction, and label that vertex $1$. Thus, we can assume every vertex is non-isolated in both $G\_1$ and $G\_2$, which is only possible if $\textbf{G}$ is an even cycle, its edges alternating between $G\_1$ and $G\_2$. It follows that $\textbf{G}$ can be properly $2$-colored.



Chromatic number of a family of graphs
The concept you introduce is called a cooperative coloring. Check out, e.g., [this paper](https://arxiv.org/pdf/1806.06267.pdf). Theorem 1 (with a reference to another paper) claims a negative answer to your question; but there is other information you may find relevant.


