Let $\{a_i\}_{i\geq 1}\subset [0,1] $ . If $f:[0,1]\longrightarrow [0,1]\backslash \{a_1,a_2,...,a_k\}$ is a bijection, then $f$ has infinite discontinuities.
Proof: For the sake of contradiction, assume that $f$ has finite discontinuities, call them $\{x_i\}_{1\leq i\leq N}$. Since the amount of $x_i$ are finite, create intervals seperated by $x_i$, call them $A_i=(x_{n_i},x_{n_i+1})$. Now since $f$ has finite discontinuous points, we define $g:[0,1]\longrightarrow [0,1]$ where $g(x)=\lim_{t\to x}f(t)$ and the image of $A_n$: $V_n:=f(A_n)=(g(x_{n_i}),g(x_{n_i+1}))$. Note that there are $k$ amount of $g(x_i)$ which are equal to $a_j$ for some $j$ . Since $x_i$ are discontinuities, we have that $g(x_i)\neq f(x_i)$ (by the definition of discontinuity). By Intermediate Value Theorem, there exists some $\eta_i \in(x_{k_i},x_{k_i+1})$ such that $f(\eta_i)\in V_j$ for some $j$ . By bijection of $f$, we cannot have that $f(x_i)\in V_j$. Therefore, $f(x_i)$ must all be on the boundaries of $V_i$. That is, it must be the case that $f(x_i)=g(x_j)$ for some $i\neq j$. But we know that there are $N$ amount of $x_i$, yet $N-k$ amount of $g(x_i)$ that $x_i$ could be mapped into by $f$. These two sets with different cardinality cannot create a bijection. Thus contradicting that $f$ is bijective. (Note that if $f$ has infinite discontinuities, then this could happen because a bijection from $\mathbb{N}$ and $\mathbb{N}$ with $k$ elements removed is possible).
An example of $f$:
$$f(x)= \begin{cases} a_{n+k} \text{ , if } x=a_n \\ x\text{ otherwise} \end{cases}$$
Is my proof correct?
