Prob. 7, Sec. 21, in Munkres' TOPOLOGY, 2nd ed: Uniform convergence and convergence in the uniform metric

Here is Prob. 7, Sec. 21, in the book Topology by James R. Munkres, 2nd edition:

Let $X$ be a set, and let $f_n \colon X \to \mathbb{R}$ 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 $f_n \colon X \to Y$ 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 $f \colon X \to Y$. And, given that $d$ is a metric on set $Y$, the function $\bar{d} \colon Y \times Y \to \mathbb{R}$ defined by $$\bar{d} (y, w) = \min \{ \ d(y, w), 1 \ \}$$ 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 $$\bar{\rho} ( f, g) \colon= \sup \left\{ \ \bar{d} ( \ f(x) \ , \ g(x) \ ) \ \colon \ x \in X \ \right\} = \sup \left\{ \ \min \{ \ d ( \ f(x) \ , \ g(x) \ ) \ , 1 \ \} \ \colon \ x \in X \ \right\}.$$ Then this function $\bar{\rho} \colon Y^X \times Y^X \to \mathbb{R}$ is a metric on $Y^X$.

Am I right?

Now we can state the following:

A sequence of functions $f_n \colon X \to Y$ converges uniformly to a function $f \colon X \to Y$ 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:

Let $\varepsilon > 0$ be given.

Suppose that the sequence of functions $f_n \colon X \to Y$ converges uniformly to a function $f \colon X \to Y$. 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 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 $x \in X$, $$\bar{d} \left( f_n (x), f(x) \right) = \min \left\{ \ d \left(\ f_n(x) \ , \ f(x) \ \right)\ , \ 1 \ \right\} < 1,$$ 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 $f_n \colon X \to Y$ converges uniformly to the function $f \colon X \to Y$.

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?

You only have to consider, for convergence purposes, $\varepsilon < 1$.
$$\forall x \in X: d(f(x),f_n(x)) \le \varepsilon \text{ iff } \bar{\rho}(f,f_n)\le \varepsilon$$