How to construct a convergent sequence of pre-images of a convergent sequence? This question is motivated by a theorem that has been widely discussed in this site:

Proposition 1. Given $X$ and $Y$ topological spaces, with $X$ a first-countable space, if $f:X\to Y$ is continuous and the sequence $(x_n)$ in $X$ converges to a limit $x$, then $f(x_n)\to f(x)$.

An interesting converse of that question would be:

If $X$ and $\mathbb{R}^{n}$ are differentiable manifolds of the same dimention, then given a proper, surjective and differentiable map $f:X\to \mathbb{R}^{n}$ and a convergent sequence  $(y_n)$ in $\mathbb{R}^{n}$, then there exist a convergent sequence $(x_n)$ so that $f(x_n) = y_n$, for all $k\in\mathbb{N}$.

At least for real-valued function, i.e., for $f: \mathbb{R} \to \mathbb{R}$, the question seems to have a positive answer. The reason for it is that proper and subjective maps explodes in both directions, but I couldn't find an actual proof.
P.S.: If needed for an easy proof, it might be assumed that the derivative has full rank in an open and dense set of $X$, but you can drop this assumption anytime you want.

Bounty question.: If $X$ and $\mathbb{R}^{n}$ are differentiable manifolds of the same dimention, then given a proper, surjective and differentiable map $f:X\to \mathbb{R}^{n}$ so that $f^{-1}(y)$ is connected. Then, for any convergent sequence  $(y_n)$ in $\mathbb{R}^{n}$, there exist a convergent sequence $(x_n)$ so that $f(x_n) = y_n$, for all $k\in\mathbb{N}$.

 A: As requested, here is an example to show that the theorem cannot hold for smooth, proper surjections between manifolds with nonempty boundary. At the very least the example shows that the theorem fails for continuous maps between locally connected, locally compact, separable metric spaces.
Let
$$X=\{(x,1)\in\mathbb{R}^2\mid x\geq0\}\cup\{(y,-1)\in\mathbb{R}^2\mid y\leq 0\}$$
carry the subspace topology inherited from $\mathbb{R}^2$. Define $f:X\rightarrow \mathbb{R}$ by
$$f(x,t)=x.$$
Then $f$ is proper and surjective. That is, $f$ is continuous and closed, and $f^{-1}(x)$ is compact for each $x\in\mathbb{R}$. In fact $f^{-1}(x)$ is a singleton whenever $x\neq 0$, while $|f^{-1}(0)|=2$.
Now consider the sequence $\{y_n=(-1)^n/n\mid n\in\mathbb{N}\}$ in $\mathbb{R}$, which converges to $0$. Letting $x_n=((-1)^n/n,(-1)^n)$ we obtain a sequence $\{x_n\mid n\in\mathbb{N}\}$ in $X$ satisfying $f(x_n)=y_n$ for each $n\in\mathbb{N}$. Of course $x_n$ is the unique point of $X$ with $f(x_n)=y_n$. The sequence $\{y_n\}$ converges in $\mathbb{R}$, but $\{x_n\}$ does not converge in $X$. Since there is no other choice for the sequence $\{x_n\}$, we see that Theorem 1 fails for the spaces and maps in the present example.
Remark 1: The example above can be altered so that both target and domain are compact and connected: simply take $f:[-1,1]\rightarrow S^1$ given by $f(x)=\exp(\pi it)$. $\quad\square$
Remark 2: If we consider maps $f:X\rightarrow Y$ which are proper and monotone (ie $f^{-1}(y)$ is compact and connected for each $y\in Y$), then it seems likely that the theorem should be true. $\quad\square$
Here is a positive result.

Lemma: Let $f:X\rightarrow Y$ be a map between Hausdorff spaces $X,Y$. Assume that  $X$ is first-countable and that $f$ is an open surjection. Then for any $y\in Y$, any $x\in f^{-1}(y)$, and any sequence $\{y_n\}_\mathbb{N}\subseteq Y$ converging to $y$, there is a sequence $\{x_n\}_\mathbb{N}\subseteq X$ which converges to $x$ and satisfies $f(x_n)=y_n$ for each $n\in\mathbb{N}$.

Proof: Given $y_n\rightarrow y$ and $x\in f^{-1}(y)$, we will produce the sequence $\{x_n\}_\mathbb{N}$. Begin by fixing a neighourhood base $\{U_n\}_\mathbb{N}$ for $x$ consisting of a sequence of decreasing open subsets of $X$. For $n\in\mathbb{N}$ write $V_n=f(U_n)$. Since $f$ is open and surjective, $\{V_n\}_\mathbb{N}$ is a neighbourhood base for $y=f(x)$ (note that $Y$ is necessarily first-countable).
Now, for each $k\in\mathbb{N}$ there is $N(k)$ such that $y_n\in V_k$ whenever $n\geq N(k)$. Fixing an $N(k)$ for each $k\in\mathbb{N}$ we can, without loss of generality, assume that $1<N(k)<N(k+1)$.
Now choose $x_n\in X$ as follows. If $n<N(1)$, then take any $x_n\in f^{-1}(y_n)$. If $N(k)\leq n<N(k+1)$ pick any $x_n\in f^{-1}(x_n)\cap U_k$.
The resulting sequence $\{x_n\}_\mathbb{N}$ lifts $\{y_n\}_\mathbb{N}$, so we'll be done if we can show that it converges to $x$. To this end suppose $W\subseteq X$ is a neighbourhood of $x$. Let $k\in\mathbb{N}$ be such that $x\in U_k\subseteq W$. Since $y_n\in V_k$ whenever $n\geq N(k)$, we have $x_n\in U_k\subseteq W$ whenever $n\geq N(K)$. This shows that $x_n\rightarrow x$. $\quad\square$
Now this can be put to use as follows. Suppose $f:X\rightarrow Y$ is a $C^1$-map between manifolds $X,Y$. The implicit function theorem implies that $f$ is open on $X\setminus D$, where $D=\{x\in X\mid rank(df)<\dim X\}$

Assume that $\dim Y>1$ and that $\{x\in X\mid rank(fd),\dim X\}$ consists of isolated points. Then $f$ is open.

The details of this are contained in the paper On Open Maps, Amer. Math. Monthly, 96 (1989), 242-243, which is authored by J. Crowe and D. Samperi.
The point is that any open map has the required sequence lifting property. There just happen to be an abundance of open maps when we impose a few conditions. Feel free to let me know if every proper $C^1$ map you would like to consider meets these conditions.
