Show that $\varphi: \overline{E} \times K \longrightarrow N$ is continuous 
Let $K$ compact in a metric space $M$, $N$ is a metric space, $\mathcal{C}(K,N)$ the space of continuous functions $f: K \longrightarrow N$ and $E \subset \mathcal{C}(K,N)$ a family of functions such that $\overline{E}$ is compact in $\mathcal{C}(K,N)$ with respect to the topology generated by the metric of uniform convergence (see the the post scriptum for the definition). Fix $a \in K$ and define $\varphi: \overline{E} \times K \longrightarrow \mathbb{R}$, $\varphi(f,x) := d(f(x),f(a))$. Show that $\varphi$ is continuous.

Initially, I was stuck because I don't know what is the metric defined on $\overline{E} \times K$ (observe that I can't take any metric since that I don't know if this space is a finite-dimensional normed space), but I tried prove the continuity of $\varphi$ via sequence and I would like to know if what I did is right.

$\textbf{My attempt:}$
Let be $(f_n,x_n)$ a sequence in $\overline{E} \times K$ such that $\lim\limits_{n \rightarrow \infty} (f_n,x_n) = (f,x)$. By continuity of $f$ and the metric $d$,
$$\lim\limits_{n \rightarrow \infty} \varphi(f_n,x_n) = \lim\limits_{n \rightarrow \infty} d(f_n(x_n),f_n(a)) = \lim\limits_{n \rightarrow \infty} d(f_n(x_n),f_n(a)) = d(\lim\limits_{n \rightarrow \infty} f_n(x_n), \lim\limits_{n \rightarrow \infty} f_n(a)) = d(\lim\limits_{n \rightarrow \infty} f_n(x_n), f(a)),$$
so the continuity of $\varphi$ will follow if I'm be able to show that $\lim\limits_{n \rightarrow \infty} f_n(x_n) = f(x)$. By the convergence of $(f_n)$ to $f$ and by the continuity of $f$ (observe that $f$ is continuous because $\overline{E}$ is compact by hypothesis, therefore is closed), we know that, given $\varepsilon > 0$, exist $N \in \mathbb{N}$ such that
$$n \geq N \Longrightarrow d(f_n(x),f(x)) < \frac{\varepsilon}{3}$$
and
$$n \geq N \Longrightarrow d(f(x_n),f(x_n)) < \frac{\varepsilon}{3}$$
Thus,
$$d(f_n(x_n),f(x)) \leq d(f_n(x_n),f_N(x_n)) + d(f_N(x_n),f_N(x)) + d(f_N(x),f(x)) < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon,$$
whenever $n \geq N$, therefore $\lim\limits_{n \rightarrow \infty} f_n(x_n) = f(x)$ and the continuity of $\varphi$ is proved. $\square$

Thanks in advance!
$\textbf{EDIT:}$
$\textbf{Definition:}$ if $X \subset M$ is arbitrary and $f,g: X \longrightarrow M$ are bounded functions, then the metric of uniform convergence is defined by
$$d(f,g) := \sup_{x \in X} d(f(x),g(x)).$$
 A: I assume, that you have a norm $\lVert \cdot\rVert_N$ on $N$; otherwise, you cannot define the sup norm  $\lVert \cdot\rVert_{\infty}$ on $\mathcal{C}(K,N)$, right?
Let $(f_n)_{n \geq 0}$ be a sequence in $\bar{E}$ and $f \in \bar{E}$ with 
$$ \lim_{n \rightarrow \infty} \lVert f_n - f \rVert_{\infty}  = 0 .$$
Furthermore, let $(x_n)_{n \geq 0}$ be a sequence in $K$ and $x \in K$ with 
$$ \lim_{n \rightarrow \infty} d_M( x_n, x)  = 0 .$$
If we can show, that this is sufficient to conclude, that 
$$ \lim_{n \rightarrow \infty} \lvert \varphi(f_n, x_n)  - \varphi(f, x) \rvert = 0 ,$$ 
then we can conclude, that $\varphi$ is continuous.
To see, that this is the case, we calculate (using the reverse triangle inequality for our first upper bound)
$$ \lvert \varphi(f_n, x_n)  - \varphi(f, x) \rvert = $$
$$ \lvert \lVert f_n(x_n) - f_n(a) \rVert_N - \lVert f(x) - f(a) \rVert_N \lvert \leq $$
$$  \lVert (f_n(x_n) - f_n(a)) - (f(x) - f(a)) \rVert_N \leq $$
$$  \lVert f_n(x_n) - f(x) \rVert_N + \lVert  f_n(a) - f(a)) \rVert_N \leq $$
$$  \lVert f_n(x_n) - f(x) \rVert_N + \lVert  f_n - f \rVert_{\infty} \leq $$
$$  \lVert f_n(x_n) - f(x_n) \rVert_N + \lVert f(x_n) - f(x) \rVert_N +  \lVert  f_n - f \rVert_{\infty} \leq $$
$$  \lVert f_n - f \rVert_{\infty} + \lVert f(x_n) - f(x) \rVert_N +  \lVert  f_n - f \rVert_{\infty}  $$ 
Taking the limit $n \rightarrow \infty$ in this chain of inequalities completes the proof.
