Let $M$ be a $C^r$ manifold $r, \geq 1$, and $A \subset M$ a connected subset .Suppose that there is a $C^r$ retraction $f: M \rightarrow A$, i.e. $f|A=$ identity. Then $A$ is a $C^r$ manifold.
I'm having trouble understanding where the retraction plays avpart. Why can't I just take charts $(\varphi_i, U_i), (\varphi_j, U_j)$ of $M$, restrict to $A$, then we have $\varphi_j\varphi_i^{-1}$ is of class $C^r$ which is a chart of $A$ thus making $A$ a $C^r$ manifold?
Recall that in the definition of a submanifold you need that $A\cap U=\phi^{-1}(\mathbb{R}^k)$ for a fixed $k\in \mathbb{N}$, and so here is where we need the retraction and connected conditions, in general you will not have that this $k$ is fixed. And also recall that you need the charts to be homeomorphisms between open sets, if $A$ is not open you might not have that $\phi(U\cap A)$ is open in $\mathbb{R}^n$ , however if $A$ is open your reasoning works and you have that all open subsets are $n-$dimensional submanifolds.