Let's suppose $X,Y \in \mathbb{R}^{\mathbb{N}}$ are random sequences.
I'd like to show that:
$$ \begin{equation} (X_1,...,X_n) \overset{d}{=} (Y_1,...,Y_n) \implies X \overset{d}{=} Y \end{equation}\tag{1}$$
where $X_i,Y_i \in \mathbb{R}$ are random variables and $\overset{d}{=}$ specifies equality in distribution.
In particular, the approach I'd like to use involves convergence of elements in $\mathbb{R}^n$ to elements in $\mathbb{R}^\mathbb{N}$:
$$ \begin{equation} \lim_{n \to \infty}(X_1,...,X_n) \overset{d}{=} X\end{equation} \tag{2}$$
At first glance this might seem silly as $\mathbb{R}^n$ and $\mathbb{R}^\mathbb{N}$ are different spaces but I think this statement can be made precise by arguing that $(2)$ is equivalent to:
$$ \begin{equation} \sum_{n=1}^{\infty} \frac{P(X_n \overset{d}{=} X_{(n)})}{2^n} = 1\end{equation} \tag{3}$$
which may be demonstrated using the definition of convergent series.
To those who are more familiar with probability theory, might there be any important gaps in my line of reasoning?
It doesn't strictly make sense because, as you say, convergence in distribution only makes sense for a sequence of random objects taking values in the same state space.
However, if you define $X^{(n)}$ as the random sequence $X^{(n)} = (X_1, X_2, \dots, X_n, 0, 0, \dots)$, then you ought to be able to show that $X^{(n)} \to X$ in distribution.
But I suspect this approach is overcomplicated. I would go for a $\pi$-$\lambda$ argument. Show that the collection $\mathcal{L}$ of Borel sets $B \subset \mathbb{R}^{\mathbb{N}}$ for which $P(X \in B) = P(Y \in B)$ is a $\lambda$-system. Now consider the collection $\mathcal{P}$ of sets $B \subset \mathbb{R}^{\mathbb{N}}$ of the form $B = A \times \mathbb{R}^{\mathbb{N}}$ where $A$ is a Borel subset of $\mathbb{R}^n$ for some $n$. Show that this is a $\pi$-system that generates the Borel $\sigma$-algebra of $\mathbb{R}^{\mathbb{N}}$. And your hypothesis (1) is precisely the statement that $\mathcal{P} \subset \mathcal{L}$, so the $\pi$-$\lambda$ theorem implies that $\mathcal{L}$ contains all Borel sets, i.e. $X \overset{d}{=} Y$.