$\newcommand{\pl}{\partial}$ $\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$
Let $\M,\N$ be smooth compact, connected, oriented manifolds of the same dimension.
Let $\,f_n:\M \to \N$ be a sequence of smooth orientation-preserving embeddings, which converges uniformly to a smooth immersion $f:\M \to \N$.
Is it true that for every $q \in \N$, the inverse image $f^{-1}(q)$ is finite?
The following example shows that $f$ does not need to be injective:
$\M=[0,2\pi],\N=\mathbb{S}^1 $, and
$$ f_n(t)=e^{i(1-\frac{1}{n})t}, f(t)=e^{ it}.$$
Then $f(0)=f(2\pi)$.
By "multiplying\tensoring" $f_n,f$ with $\text{Id}_{\mathbb{S}^1}$ one can see that in general there is no finite subset of $\M$ that removing it will make $f$ injective.
Any smooth immersion $f : \mathcal{M} \to \mathcal{N}$ between compact manifolds has the property that $f^{-1}(q)$ is finite for all $q \in \mathcal{N}$. This is true regardless of whether $f$ arose from some limiting process, and regardless of whether $\mathcal{M}$ and $\mathcal{N}$ have the same dimension.
For the proof, every immersion is a local embedding, meaning that for each $x \in \mathcal{M}$ there is an open neighborhood $U \subset \mathcal{M}$ of $x$ such that $f | U$ is injective. It follows that for each $q \in \mathcal{N}$, the subset $f^{-1}(q) \subset \mathcal{M}$ is discrete. Also, $f^{-1}(q)$ is a closed subset of $\mathcal{M}$. But every discrete, closed subset of a compact space is finite.