Let $(x_n)_{n=1}^{\infty}$ be $1-1$ if $\left(\forall n\in \mathbb{N}\right) i\ne j\implies x_i\ne x_j$ and let $f:\mathbb R\to \mathbb R$ and $a$ an accumulation point on the domain of $f$.
Prove: $\lim_{x\to a}f(x)=b \iff$ for any $1-1$ sequence $x_n$ s.t $x_n\to a$ $f(x_n)\to b$
$\Rightarrow:$ by definition $\lim_{x\to a}f(x)=b$ if for all $x_n\to a$ we have $f(x_n)\to b$ in particular that is also true if $x_n$ is $1-1$
$\Leftarrow:$ We know that $f(x_n)\to b$ for all $1-1$ sequences $x_n$ we have to show it is true for non $1-1$ sequences such that $x_n\to a$:
case a: the sequence is the constant $a$, therefore $f(a)\to b$
case b: there are finite elements of $x_n$ which are not $1-1$, we can remove them and still $x_n\to a$ and $f(a)\to b$
case c: there are infinite elements of $x_n$ which are not $1-1$, the sequence converges to $a$ so infinite elements of the sequences is $\varepsilon$ close to $a$ and therefore $f(x_n)\to b$
Is the proof correct?
P.S another try of $\Leftarrow:$ Proof by contradiction assume that any $1-1$ sequence $x_n\to a$ but $lim_{x\to a}f(x)\neq b$, that mean that there are $2$ sequence $x_n\to a$ and $z_n\to a$ which are $1-1$ such that $b_1\leftarrow f(x_n)\neq f(z_n)\to b_2 $ contradiction to the assumption that for all $1-1$ we have $f(x_n)\to b$
Suppose $\lim_{x \to a} f(x) = b$. Then for any sequence $x_n \to a$ such that $x_n = a$ for at most a finite number of $n$ then $f(x_n) \to b$. If the sequence is injective then $x_n$ can equal $a$ at most once and so $f(x_n) \to b$.
For the other direction, the contrapositive is a little easier, I believe:
Suppose there is some $\epsilon>0$ such that for all $\delta>0$ there is some $x$ satisfying $0<|x-a| < \delta$ such that $|f(x)-b| \ge \epsilon$. Choose $\delta =1$ to start to get $x_1$. Now suppose we have chosen $x_1,..., x_n$, then let $\delta = \min_{k \in \{1,...,n\}}({1 \over n+1}, |x_k-a|)$ and pick $x_{n+1}$. Then $x_n \to a$ and $x_n$ is injective but $f(x_n) \not\to b$.