Here is Prob. 7, Sec. 21, in the book Topology by James R. Munkres, 2nd edition:
Let $X$ be a set, and let $f_n \colon X \longrightarrow \mathbb{R}$ be a sequence of functions. Let $\bar{\rho}$ be the uniform metric on the space $\mathbb{R}^X$. Show that the sequence $\left( f_n \right)$ converges uniformly to the function $f \colon X \to \mathbb{R}$ if and only if the sequence $\left( f_n \right)$ converges to $f$ as elements of the metric space $\left( \mathbb{R}^X, \bar{\rho}\right)$.
Here is the definition in Munkres of a uniformly convergent sequences of functions:
Let $f_n \colon X \longrightarrow Y$ be a sequence of functions from the set $X$ to the metric space $Y$. Let $d$ be the metric for $Y$. We say that the sequence $\left( f_n \right)$ converges uniformly to the function $f \colon X \to Y$ if given $\varepsilon > 0$, there exists an integer $N$ such that $$ d \left( f_n (x), f(x) \right) < \varepsilon $$ for all $n > N$ and all $x$ in $X$.
My Attempt:
Let $X$ be a (non-empty) set, and let $(Y, d)$ be a metric space. Let $Y^X$ denote the set of all the functions $f \colon X \longrightarrow Y$. And, given that $d$ is a metric on set $Y$, the function $\bar{d} \colon Y \times Y \longrightarrow \mathbb{R}$ defined by $$ \bar{d} (y, w) \colon= \min \big\{ d(y, w), 1 \big\} $$ for all $y, w \in Y$ is also a metric on $Y$, as has been shown by Munkres in Theorem 20.1.
Now for any elements $f, g \in Y^X$, let $$ \begin{align} \bar{\rho} ( f, g) &\colon= \sup \left\{ \ \bar{d} \big( f(x) \, , \, g(x) \big) \colon x \in X \ \right\} \\ &= \sup \left\{ \, \min \big\{ d \big( \ f(x) , g(x) \big) \, , 1 \big\} \colon x \in X \right\}. \end{align} $$ Then this function $\bar{\rho} \colon Y^X \times Y^X \longrightarrow \mathbb{R}$ is a metric on $Y^X$.
Am I right?
Now we can state the following:
A sequence of functions $f_n \colon X \longrightarrow Y$ converges uniformly to a function $f \colon X \longrightarrow Y$ if and only if the sequence $\left( f_n \right)$ converges in the metric space $\left( Y^X, \bar{\rho} \right)$ to the element $f \in Y^X$.
Am I right?
Proof:
Suppose that the sequence of functions $f_n \colon X \longrightarrow Y$ converges uniformly to a function $f \colon X \longrightarrow Y$. Let $\varepsilon > 0$ be given. Then there exists a natural number $N$ such that $$ d \left( f_n(x), f(x) \right) < \frac{\varepsilon}{2} $$ for all $n > N$ and all $x \in X$. So $$ \bar{d} \left( f_n(x), f(x) \right) \leq d \left( f_n(x), f(x) \right) < \frac{\varepsilon}{2} $$ for all $n > N$ and all $x \in X$. Therefore, $$ \bar{\rho} \left( f_n, f \right) = \sup \left\{ \bar{d} \left( f_n(x), f(x) \right) \, \colon \, x \in X \ \right\} \leq \frac{\varepsilon}{2} < \varepsilon $$ for all $n > N$. Thus it follows that the sequence $\left( f_n \right)_{n \in \mathbb{N} }$ converges to $f$ in the metric space $\left( Y^X, \bar{\rho} \right)$.
Am I right?
Conversely, suppose that the sequence $\left( f_n \right)_{n \in \mathbb{N} }$ converges to $f$ in the metric space $\left( Y^X, \bar{\rho} \right)$. Then, given $\varepsilon > 0$, there exists a natural number $N$ such that $$ \bar{\rho} \left( f_n, f \right) < \frac{\varepsilon}{ 1 + \varepsilon} $$ for all $n > N$. But $$ \bar{d} \left( f_n(x), f(x) \right) \leq \bar{\rho} \left( f_n, f \right) $$ for all $x \in X$. So we can conclude that $$ \bar{d} \left( f_n(x), f(x) \right) < \frac{\varepsilon}{ 1 + \varepsilon} \tag{1} $$ for all $n > N$ and for all $x \in X$.
But as $\varepsilon > 0$, so $$ 0 < \frac{\varepsilon}{1+ \varepsilon} < 1, $$ and thus (1) implies that for all $n > N$ and all $x \in X$, we have $$ \bar{d} \left( f_n (x), f(x) \right) = \min \left\{ d \left(f_n(x) , f(x) \right) \, , \, 1 \right\} < 1, $$ and so $$ \bar{d} \left( f_n (x), f(x) \right) = d \left( f_n(x) , f(x) \right) \tag{2} $$ for all $n > N$ and for all $x \in X$.
Now using (2) in (1) we can conclude that $$ d \left( f_n(x) , f(x) \right) < \frac{\varepsilon}{1 + \varepsilon} < \varepsilon $$ for all $n > N$ and all $x \in X$. Thus it follows that the sequence of functions $f_n \colon X \longrightarrow Y$ converges uniformly to the function $f \colon X \longrightarrow Y$.
Am I right?
Is every part of the above proof correct? If so, then is my presentation accessible enough, especially for a student at an elementary level?
Or, are there any problems?
You only have to consider, for convergence purposes, $\varepsilon < 1$.
And in that case
$$\forall x \in X: d(f(x),f_n(x)) \le \varepsilon \text{ iff } \bar{\rho}(f,f_n)\le \varepsilon$$
That is really all that matters.