A complex sequence of functions $(f_n)$ is continuously convergent iff it's compactly convergent against a continuous function Let $G \subseteq \mathbb{C}$ be a region in $\mathbb{C}$, i.e. $G$ is open, nonempty and connected, and let $f_n: G \to \mathbb{C}$ be a sequence of complex-valued functions, with $n \in \mathbb{N}$. We call $(f_n)$ continuously convergent iff $(f_n(x_n))_{n \in \mathbb{N}} \subseteq \mathbb{C}$ is a convergent sequence, for all convergent $(x_n) \subseteq G$. I now want to show:

$(f_n)$ converges compactly (i.e. converges uniformly on each compact
  $K \subseteq G$) to a continuous function $f: G \to \mathbb{C}$
  iff $(f_n)$ converges continuously.

For "=>", I thought about the following: if $(f_n)$ converges uniformly on each compact subset against a continuous $f$, then we have that $\lim_{n \to \infty} f_n(x) = f(x)$; and since $f$ is continuous, we also have $\lim_{n \to \infty} f(x_n) = f(x)$ for every sequence with $x_n \to x$ (in $\mathbb{C}$). Can I already combine that to $f(x) = \lim_{n \to \infty} \left( \lim_{m \to \infty} f_n(x_m)    \right) = \lim_{n \to \infty} f_n(x_n)$ to get continuous convergence?
I'm not really sure how to approach "<=", though. If $f(x) = \lim_{n \to \infty} f_n(x_n)$, then I would need to show that $f$ converges uniformly on each compact subset, i.e. that $\lim_{n \to \infty} \sup_{x \in K} (f_n(x) - f(x)) = 0$, which is already a way harder claim than pointwise convergence, and I don't know how I can get there with the definition of $f$ being continuously convergent. (It's obvious to me that $f_n$ is pointwise convergent, since we get the very definition of pointwise convergence if we simply set $x_n = x$.)
(I'm also told that this result holds for general metric spaces, not just in $\mathbb{C}$, but for me, it's sufficient to prove it in $\mathbb{C}$ for now. I'm assuming that the solutions for both directions aren't to hard to generalize when having solved  them for the complex case.)
 A: Proof of $\Longrightarrow$: Let $(f_n)$ be a sequence of functions in $G$
which converges to a continuous function $f$, uniformly on each compact
subset of $G$. Let $(x_n)$ be a convergent sequence in $G$, $x_n \to x \in G$. We have to show that $(f_n(x_n))$ is convergent.
Choose a compact disk $K$ such that $x \in K \subset G$. Then $x_n \in K$
for $x \ge n_0$, and
$$
 \lvert f_n(x_n) - f(x) \rvert \le  \lvert f_n(x_n) \rvert
 - f(x_n)) + \lvert f(x_n) - f(x) \rvert \, .
$$
The first term converges to zero because  $f_n \to f$ uniformly on $K$,
and the second term converges to zero because $f$ is continuous in $x$.
It follows that $f_n(x_n) \to f(x)$.
Proof of $\Longleftarrow $:
Let $(f_n)$ be a sequence of functions in $G$ which is "continuously
convergent," i.e.

$$ \tag{*}
  (x_n) \text{ is convergent in } G \Longrightarrow
 (f_n(x_n)) \text{ is convergent in } \Bbb C \, .
$$

As you already noticed, $(*)$ implies that $(f_n)$ is pointwise
convergent, and we can define 
$$
  f: G \to \Bbb C, \, f(x) := \lim_{n \to \infty} f_n(x) \, .
$$
We have to show that $f_n \to f$ locally uniformly in $G$,
and that is done in three steps.

Claim 1: For any convergent sequence $x_n \to x \in G$,
  $f_n(x_n) \to f(x)$ holds.

Proof of Claim 1: Consider the sequence
$$
   (y_n)=  (x_1, x, x_3, x, x_5, x, \ldots) \, .
$$
From $y_n \to x$ is follows that $f_n(y_n)$ is convergent.
Since $f_n(y_n) = f(x)$ for all even $n$, the limit must be $f(x)$
and therefore the subsequence of $f_n(x_n)$ with odd indices
converges to $f(x)$. Then consider
$$
   (z_n)=  (x, x_2, x, x_4, x, x_6, \ldots) \, .
$$
to show that  the subsequence of $f_n(x_n)$ with even indices
converges to $f(x)$ as well. 

Claim 2: The limit function  $f$ is continuous.

This follows from Claim 1, and various proofs can be found in


*

*If $f_n(x_n) \to f(x)$ whenever $x_n \to x$, show that $f$ is continuous.



Claim 3: $f_n \to f$ uniformly on each compact set $K \subset G$.

Proof of Claim 3: Assume that this is false. Then there exists
a compact set $K \subset G$, an $\varepsilon > 0$, a subsequence
$(f_{n_k})$ of $(f_n)$, and a sequence $(x_k)$ in $K$ such that
$$
  \lvert f_{n_k}(x_k) - f(x_k) \rvert > \varepsilon \text{ for all } k \, . 
$$
Since $K$ is compact, $(x_k)$ has a convergent subsequence.
Without loss of generality we can assume that $(x_k)$ is convergent,
$x_k \to x \in K$. Now
$$
 \varepsilon < \lvert f_{n_k}(x_k) - f(x_k) \rvert \
   \le \lvert f_{n_k}(x_k) - f(x) \rvert + \lvert f(x) - f(x_k) \rvert
$$ 
The first term converges to zero because of Claim 1, 
and the second term converges to zero because of Claim 2 ($f$
is continuous in $x$). This is a contradiction.
