Proving that a limit exists given information about converging sequences. 
Let $(Z,d), (Y,d')$ be metric spaces and $f: X \subset Z \to Y$ a
  function. Let $a$ be a limit point of $X$ in $Z$. Define the function 
$$\hat{f}_b: X \cup\{a\} \to Y: x \mapsto \begin{cases}{f(x) \quad x
 \neq a} \\ b \quad\quad x = a \end{cases}$$
where $b\in Y$ is a fixed element. Given the following statements:
(1) For any two sequences $(x_n)_n, (y_n)_n$ in $X -\{a\}$ such that
   $x_n \to a$ and $y_n \to a$, we have: $(f(x_n))_n, (f(y_n))_n$ are
  convergent sequences and share the same limit.
(2) $\lim_{x \to a} f(x)$ exists
Prove that $(1) \implies (2)$

My attempt:
Call $b$ the unique limit of the image of all sequences that consist of elements in $X - \{a\}$ and converge to $a$.
Since we have that $\lim_{x \to a} f(x) = b \iff \hat{f}_b$ is continuous in $a$, it suffices to show that $\hat{f}_b$ is continuous in $a$.
To do this, we take an arbitrary sequence $(x_n)_n$ such that $x_n \to a$. We must prove that $\hat{f}_b(x_n) \to \hat{f}_b(a) = b$.
If the sequence $(x_n)_n$ contains only a finite amount of terms that are $a$, the sequence $f(x_n)$ converges to $b$.
However, I have trouble proving that $f(x_n) \to b$ if $(x_n)$ contains the term $a$ an infinite amount of times.
Please, post only solutions that prove that $\hat{f}_b$ is continuous at $a$
 A: Actually, you have to prove that $\hat{f}_b(x_n) \xrightarrow{n\to\infty} b$, since $a$ is not in the domain of $f$.
Let $(x_n)_{n=1}^\infty$ be a sequence in $X \cup \{a\}$ such that $x_n \xrightarrow{n\to\infty} a$.
If $a$ occurs only finitely many times in $(x_n)_{n=1}^\infty$ (let $n_0$ be the last occurrence), then the sequence $(x_n)_{n=n_0+1}^\infty$ is in $X \setminus \{a\}$ so $(f(x_n))_{n=n_0+1}^\infty$ converges to $b$. Then  $(f(x_n))_{n=1}^\infty$ also converges to $b$.
Assume that $a$ occurs infinitely many times in $(x_n)_{n=1}^\infty$. Let $(x_{q(n)})_{n=1}^\infty$ be the subsequence of $(x_n)_{n=1}^\infty$ containing all elements not equal to $a$.
Using your assumption, $\left(\hat{f}_b(x_{q(n)})\right)_{n=1}^\infty = (f(x_{q(n)}))_{n=1}^\infty$ converges to $b$.
Let's show that $\left(\hat{f}_b(x_n)\right)_{n=1}^\infty$ converges to $b$. Let $\varepsilon > 0$. Since $f(x_{q(n)}) \xrightarrow{n\to\infty} b$, there exists $n_0 \in \mathbb{N}$ such that $n \ge n_0 \implies d'(f(x_{q(n)}), b) < \varepsilon$.
Then, for $n \ge q(n_0)$ we have that $d'\left(\hat{f}_b(x_{n}), b\right) < \varepsilon$, since $d'\left(\hat{f}_b(x_{n}), b\right)$ is either $d'\left(f(x_{q(n)}), b\right)$ for some $n \ge n_0$, or it is $d'\left(\hat{f}_b(a), b\right) = d'(b,b) = 0$.
We conclude that $\hat{f}_b(x_n) \xrightarrow{n\to\infty} b$.
