Show there is a bijection between $R^n\ and\ R^N$ Let $N=\{1,2,3,...,n\}$, show there is a bijection between $R^n\ and\ R^N$

Define $f: R^n \rightarrow R^N$ by $f(x_1,..,x_n)=g_{(x_1,...x_n)}$ where the n-tuple is an element of $R^n$ and $g_{(x_1,...x_n)}:N \rightarrow \mathbb{R}$ is defined by:

$g_{(x_1,...x_n)}(i)=x_{i}$ for $i \in N$

What I want is g to map the elements in N to the corresponding real number in the n-tuple. But I'm not sure that I can define g in this way.

But if I can here is my proof this is injective.

Suppose $(x_1,...x_n) \neq (y_1,...y_n)$ then there is at least one $i \in N$ such that $x_i \neq y_i$.

Then the function $g_{(x_1,...x_n)}(i)=x_i \neq y_i=g_{(y_1,...y_n)}$ for some $i \in N$

Since $f(x_1,..,x_n)=g_{(x_1,...x_n)}$ then $f(x_1,...,x_n)\neq f(y_1,...,y_n)$

My issue and why I think this doesn't work is because when I want to prove this is surjective, I want to prove that for any g, there is an n-tuple s.t $g=g_{(x_1,...,x_n)}$ but the way I've defined g seems to make it depend upon this tuple already existing.
You don't have to worry too much about the tuple existing. You're looking at $\mathbb{R}^n$, which is the set of all $n$-tuples with real numbers in every entry. You are completely free to choose which real numbers to put in all $n$ entries.

If you look at the map $g \mapsto (g(1), g(2), \ldots, g(n))$, then this is a well-defined map from $\mathbb{R}^N$ to $\mathbb{R}^n$, and I think you'll find that it's the inverse of $f$.
I would just appeal to cardinal arithmetic.  We have $$\mathfrak c = |\Bbb R|=2^{\aleph_0}=2^{|\Bbb N|}\\
\mathfrak c = |\Bbb R|\le |\Bbb R^n|\le |\Bbb R^{|\Bbb N|}|=(2^{|\Bbb N|})^{|\Bbb N|}=2^{|\Bbb N|{|\Bbb N|}}=2^{|\Bbb N|}=\mathfrak c$$