Let $M,N$ be smooth two-dimensional connected, oriented, compact Riemannian manifolds with boundaries, and let $f:M \to N$ be smooth and injective.
Let $f_t:M \to N$ be a smooth variation of $f$. Is $f_t$ injective for sufficiently small $t$?
If that matters, I am fine with assuming that $\det(df)>0$ and $f(\partial M) \subseteq \partial N$. (although I am not sure if it's needed.)
Here is a naive attempt: Suppose that this is not so; then we have $t_n \to 0$, $x_n \neq y_n \in M$ such that $f_{t_n}(x_n)=f_{t_n}(y_n)$. Since $M$ is compact we have $x_n \to x,y_n \to y$ (modulue subsequences). Thus $f(x)=f(y)$. Since $f$ is injective, this forces $x=y$.
So, $x_n \neq y_n, x_n,y_n \to x$ and $f_{t_n}(x_n)=f_{t_n}(y_n)$. Now, if $x \in N^o$, then $f(x) \in M^o$ (since $df_x$ is non-singular), so by the inverse function theorem, applied to $f_t|_{M^o}:M^o \to N^o$ at $x$, $f_t$ is injective in a neighbourhood of $x$. This might be contradictory with $x_n,y_n \to x$, if we can quantify the size of the neighbourhood where the IFT gives us injectivity, independently of $t$ for small $t$. (see e.g. here). A similar argument should work when $x \in \partial M$.
Can this approach work? Are there other approaches? Or is there a counter-example?
There is a standard trick for dealing with this. If your parameter $t$ is in some interval $I$, consider $F\colon M\times I\to N\times I$ given by $$F(x,t) = (f_t(x),t).$$ Assuming that $f=f_0$ is an immersion, we see that $F$ is an immersion at $(x,0)$ for all $x\in M$, and hence one-to-one in a neighborhood of $(x,0)$. Therefore, for large enough $n$, $F(x_n,t_n)=F(y_n,t_n)$ forces $x_n=y_n$, completing your contradiction.