Is a submodule of a flat module flat?

Let $A=\mathbf{Z}/4\mathbf{Z}$. Then $A$ is flat over itself, but the ideal $I:=2\mathbf{Z}/4\mathbf{Z}$ is not flat over $A$. This is because when we tensor the injection $I\hookrightarrow A$ with $I$, we get the map $I\otimes_AI\rightarrow I$ which sends $r\otimes s$ to $rs$, is visibly the zero map. The group $I\otimes_AI$ is not zero because it is isomorphic to $(\mathbf{Z}/2\mathbf{Z})\otimes_{\mathbf{Z}/2\mathbf{Z}}(\mathbf{Z}/2\mathbf{Z})\cong \mathbf{Z}/2\mathbf{Z}$. So $I\otimes_AI\rightarrow I$ is not injective.

However, if $A$ is a principal ideal domain, or more generally a Dedekind domain, then submodules of flat $A$-modules are flat, because for such a ring $A$, flat$=$torsion-free, and it is clear that submodules of torsion-free modules are torsion-free (over any domain).

Let $A$ be a ring. Then $A$ is of course flat over $A$. But sub-modules of $A$ are just ideals, and these are rarely flat over $A$.

The property that submodules of flat modules are flat is equivalent to being locally a valuation domain (or in homological terms, having weak global dimension at most 1), see https://stacks.math.columbia.edu/tag/092S for reference. To expand on The L's answer, this is in turn equivalent to having flat ideals. Since being reduced is a local property, this shows that any ring with nilpotents will serve as a counter example to your question, and in fact will have non-flat principal ideals.