Here, I tried to write the answer as explicit as possible. In fact, Noah Schweber already mentioned Lusin's example of analytic non-Borel set in the deleted answer. Also, Gio67 mentioned the Descriptive Set Theory notes. Thus, the answer is already there in Gio67's answer. I tried explaining the parts not covered by those answers.
Lemma
The set $\mathbb{N}^{\mathbb{N}}$ of sequences of natural numbers is homeomorphic to the set of irrational numbers $I$ in $[0,1]$.
This is by continued frantion expansion of irrational numbers.
Definition
Denote by $E$ the set of irrational numbers $\alpha\in [0,1]$ with the continued fraction expansion $[0;a_1,a_2,a_3\ldots]$ such that there exists a subsequence $\{n_k\}\subseteq \mathbb{N}$ with
$$a_{n_k}|a_{n_{k+1}} \ \mathrm{for} \ k\geq 1. $$
Theorem
There is a surjective continuous function from $\mathbb{N}^{\mathbb{N}}$ to $E$.
Proof.
Let $\{x_n\}\in \mathbb{N}^{\mathbb{N}}$. Construct an increasing subsequence $\{n_k\}$ by using $\{x_{3k+1}\}$ as follows.
\begin{align*}
n_1&=x_1,\\
n_{k+1}&=n_k+x_{3k+1} \ \mathrm{for}\ k\geq 1.
\end{align*}
Then we place the numbers from $\{x_{3k+2}\}$ in the following way.
\begin{align*}
a_{n_1}&=x_2,\\
a_{n_{k+1}}&=a_{n_k}x_{3k+2}\ \mathrm{for}\ k\geq 1.
\end{align*}
Enumerate remaining numbers $\mathbb{N}-\{n_k\}=\{m_k\}$ in increasing order. Then use $\{x_{3k}\}$ to fill up these partial quotients.
$$
a_{m_k}=x_{3k}\ \mathrm{for}\ k\geq 1.
$$
Then define $f(\{x_n\})=[0;a_1,a_2,a_3,\ldots]$.
Now, we have an answer to the question.
There is a $G_{\delta}$ subset $A$ of $[0,1]$ such that there is a continuous function $f:[0,1]\rightarrow [0,1]$ with $f(A)=E$.
Proof.
Consider the following composition.
$$
g:I\rightarrow \mathbb{N}^{\mathbb{N}}\rightarrow E.
$$
Then $G=\mathrm{Graph}(g)=\{(x,g(x))|x\in I\}$ forms a closed subset of a $G_{\delta}$ set $I\times [0,1]$. Then $G$ itself is a $G_{\delta}$ subset of $[0,1]^2$. Then the projection $\pi_2$ onto the second coordinate yields $\pi_2(G)=E$.
Let $\Phi:[0,1]\rightarrow [0,1]^2$ be the continuous space-filling curve. Then take the $G_{\delta}$ set $A=\Phi^{-1}(G)\subset [0,1]$. This gives $f(A)=E$ where $f=\pi_2\circ \Phi$.
$\bullet$ $\mathbb R^n$ is second countable with a countable basis $B$ given by all balls with rational radii and centers.
$\bullet$ Subspaces of second countable spaces are second countable: If $Y\subseteq X$ and $B$ is a countable basis for $X$, then $\{U\cap Y: U\in B\}$ is a countable basis for $Y$.
$\bullet$ Second countable spaces are Lindelöf: Any open cover admits a countable subcover (see for example here or here).
So an embedded submanifold $M$ of $ \mathbb R^n$ is is "countably locally" the graph of a continous function, which finishes your proof.
Best Answer
Negative example.
Let $T \subseteq [0,1]\times[0,1]$ be a Borel set with non-Borel projection $\pi_1(T)$ on the first coordinate. Let $S = [0,1]\times[0,1]$ be the quare. Our sets are made up of vertical translates of these two: $$ A = T \cup (S + (0,2)), \\ B = T \cup (S + (0,5)), $$ so that $\pi_1(A) = [0,1]$ is Borel, $\pi_1(B) = [0,1]$ is Borel, but $\pi_1(A\cap B) = \pi_1(T)$ is not Borel.