functional definition of tuples (Terrence Tao Analysis, execise 3.5.2)

I came across this particular exercise in T. Tao Analysis book:

Suppose we define an ordered $n$-tuple to be a surjective function $x : \{i ∈ N : 1 ≤ i \leq n\} \to X$ whose range is some arbitrary set $X$ (so different ordered $n$-tuples are allowed to have different ranges); we then write $x_i$ for $x(i)$, and also write $x$ as $(x_{i})_{1≤i≤n}$. Using this definition, verify that we have $(x_i)_{1≤i≤n} = (y_i)_{1≤i≤n}$ if and only if $x_{i} = y_{i}$ for all $1 ≤ i ≤ n$. Also, show that if $(X_i)_{1≤i≤n}$ are an ordered $n$-tuple of sets, then the Cartesian product, as defined in Definition 3.4.3 (Ordered n-tuple and n-fold Cartesian product), is indeed a set. (Hint: use Exercise 3.4.7 and the axiom of specification.

I started proving the first statement without looking at the hint and this was my attempt:

First, we verify that $(x_i)_{1≤i≤n} = (y_i)_{1≤i≤n}$ if and only if $x_{i} = y_{i}$ for all $1 ≤ i ≤ n$.

1. we show that $(x_i)_{1≤i≤n} = (y_i)_{1≤i≤n}\implies$$x_{i} = y_{i}$ for all $1 ≤ i ≤ n$.

Suppose $(x_i)_{1≤i≤n} = (y_i)_{1≤i≤n}$.

Let $y$ and $x$ be any two surjective functions $x:\{ 1,2,\dots,n \}\to X$ and $y:\{ 1,2,\dots,n \}\to Y$.

Thus, $(x_i)_{1≤i≤n} = (y_i)_{1≤i≤n}$ means that $x=y$, so for each index $i\in \{ 1,2,\dots n \}$ the corresponding elements in the tuples, namely, $x(i)$ and $y(i)$ are equal.

Thus, by definition, $x_i=y_i$ for all $1\leq i\leq n$.

2. we show that $x_{i} = y_{i}$ for all $1 ≤ i ≤ n\implies$$(x_i)_{1≤i≤n} = (y_i)_{1≤i≤n}$.

If $x_i=y_i$ for all $1\leq i\leq n$, we have that $x(i)=y(i)$ for all $1\leq i\leq n$.

Since the domains of $x$ and $y$ are the same and for every $i$ in the domain we have $x(i)=y(i)$, we need only show that $X=Y$ to conclude that $x=y$.

Let $a\in X$. Since $x$ is surjective, there exists $i\in \{ 1,2,\dots,n \}$ such that $a=x(i)$. Since $x(i)=y(i)$ for all $1\leq i\leq n$, we have that $a\in Y$. Thus, $X\subseteq Y$.

Similarly for $b\in Y$, since $y$ is surjective there exists $i\in \{ 1,2,\dots,n \}$ such that $b=y(i)$. Since $y(i)=x(i)$ for all $1\leq i\leq n$, we have that $b\in X$. Thus $Y\subseteq X$.

Since $Y\subseteq X$ and $X\subseteq Y$, it follows that $X=Y$.

Thus, we have that the domain and ranges of $x$ and $y$ are the same, and for every $i$ in the domain, $x(i)=y(i)$. By definition of equality for functions, $x=y$. Thus, by the defition provided by the exercise, $(x_i)_{1≤i≤n} = (y_i)_{1≤i≤n}$.

After finishing the first part of the proof, I realized there was a hint and that I never used any of the hinted propositions, so my question are:

1. Are the hints strictly used in proving the second part of the exercise?
2. If the hint was intended for the first part, is my proof at all correct (since I never used the hints, I think there is a high chance that I assumed something that I could not)
3. Also, my first instinct was to use induction on $n$, but after several attempts at this, I gave up. Should I have kept in this line of reasoning?