Let $R$ be a commutative Noetherian ring with non-zero identity and $M$, $M'$ and $M''$ be $R$-modules (not necessarily finitely generated) with $\operatorname{Supp} M \subseteq \operatorname{Supp} M'\cup \operatorname{Supp} M''$. Is it true that $\dim M \leq \max\{\dim M' , \dim M''\}$?
The Krull dimension of a module $M$ is the supremum of the non-negative integers $n$ such that there exists a chain of prime ideals in $\operatorname{Supp}M$ of length $n$.
Thanks.
Since in my first answer I assumed that all modules are finitely generated, which was not the intention of OP (who actually haven't put his question into precise form at that moment), I decided to edit my answer so that it also covers the general case.
Suppose that dimension of a module $M$ is the supremum of lengths of all chains of prime ideals contained in $\mathrm{supp}(M)$. Note that this definition agrees with the usual one in the case when module is finitely generated. Then the inequality in question holds. This is due to the following fact:
proof: We have: $$M_{\mathfrak{q}}\otimes_{R_{\mathfrak{q}}}R_{\mathfrak{p}}=M_{\mathfrak{p}}.$$ Now if $M_{\mathfrak{p}}\neq 0$, then also $M_{\mathfrak{q}}\neq 0$.
Now let $\mathrm{supp}(M)\subseteq \mathrm{supp}(M')\cup \mathrm{supp}(M'')$. Consider a chain: $$\mathfrak{p}_0\subseteq \mathfrak{p}_1\subseteq \cdots\subseteq \mathfrak{p}_n$$ of prime ideals in $\mathrm{supp}(M)$. Then we have either $\mathfrak{p}_0\in \mathrm{supp}(M')$ or $\mathfrak{p}_0\in \mathrm{supp}(M'')$ and thus by the fact, we derive that the whole chain is contained either in $\mathrm{supp}(M'')$ or in $\mathrm{supp}(M'')$. Hence: $$n\leq \max\{\mathrm{dim}(M'),\mathrm{dim}(M'')\}.$$