Is every sequence defined on $\mathbb{R}$ just a countable subset of $\mathbb{R}$?

Let $a : \mathbb{N} \rightarrow \mathbb{R}$ be some sequence. For example, $a_n$ could be $$ a_n = \frac{1}{n} $$ Can the image of every possible function from $\mathbb{N}$ to $\mathbb{R}$ be treated the same as a countable (finite or infinite) subset of $\mathbb{R}$?

It is true that every non-empty countable subset of $\mathbb R$ is the image of some function $a:\mathbb N\to\mathbb R.$ This is true if we replace $\mathbb R$ with any set $X.$

But this does not mean that a sequence is “just” a countable subset, as you wrote in your title. Sequences have an order to them, and can have repeated elements. If $a_1=0,a_{n}=1$ for $n>1,$ the countable set is $\{0,1\}.$ If $b_n=1-a_n,$ the underlying set is the same, but the sequence is different.

If we restrict to $a_n$ with no repetitions, we still have properties of sequences which depend on more than the image of $a.$ For example, $a_n=\frac{(-1)^n}n,$ then $$\lim_{n\to\infty}\sum_{k=1}^n a_k=\ln 2,$$ but there are sequences $b_n$ with the same underlying set which do not converge, or converges to another value.

At heart, a function has far more to it than it’s image.

The range of $a$ is a subset of $\mathbb R$, and is clearly countable. So the sequence $a$ clearly has a countable subset of $\mathbb R$. But as @Thomas Andrews pointed out, this subset lacks critical information, namely the order and duplicates. So the subset cannot in any meaningful way be considered equivilant to the sequence-generating function.