Prove that $\lim_{y \to x-}f(y) = a$ if for every sequence $(x_n)_n$ in $\mathbb{R}$ that increases towards x, the sequence $f(x_n)$ converges to $a$ 
Let $f: \mathbb{R} \to \mathbb{R}$ be a function. 
Prove that: $$\forall (x_n)_n \subseteq \mathbb{R}: (x_n \to x \land
 \forall n: x_n \leq x_{n+1} \implies f(x_n) \to a)$$ implies that
   $\lim_{y \to x-}f(y) = a$

My attempt:
Suppose $\lim_{y\to x-}f(y) \neq a$. This means,
$$\exists \epsilon >0: \forall \delta >0: \exists y \in \mathbb{R}: 0 < x-y < \delta \land |f(y) - a | \geq \epsilon$$
Choosing this $\epsilon > 0$ and letting $\delta := 1/n$, we can find a sequence $(y_n)_n$ in the reals such that $0 < x-y_n< 1/n$ and $|f(y_n)-a| \geq \epsilon$
This implies that $y_n \to x$ and $f(y_n) \not\to a$ and it is also clear that $(y_n)_n$ is increasing. The result follows by contraposition.
Is this correct?  

Attempt to fix it , correct me if wrong:
You have:
1)$0<x-y_n \lt 1/n$  and 
2)$|f(y_n)-a| \ge \epsilon$.
1) Implies $\lim_{ n \rightarrow \infty} y_n =a.$
You need to choose an increasing subsequence such that
$y_{n_k} \le y_{n_{k+1}} $.
Inductively:
0) Start: $x-y_0\lt 1/n_0.$
1) Choose $n_1$ such that : 
$1/n_1 \lt (x-y_0).$
Step:
2) Choose $n_{k+1}$ such that : 
$1/n_{k+1} \lt (x-y_{n_k})$.
Now apply your argument for the increasing subsequence $y_{n_{k}}.$
