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. $$\square$$

$$\textbf{EDIT:}$$

$$\textbf{Definition:}$$ if $$X \subset M$$ is arbitrary and $$f,g: X \longrightarrow M$$ are bounded functions, then the metric of uniform convergence is defined by

$$d(f,g) := \sup_{x \in X} d(f(x),g(x)).$$

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.
• Sorry, it was a typo, $\mathcal{C}(K,N)$ is not equipped with the sup norm, I edited my post and put the definition of the metric of uniform convergence – George Apr 22 at 23:11