Propositon: Given that the real sequences $\{a_n\}$ and $\{b_n\}$ are convergent, and that $\{a_n - b_n \}$ is a null sequence, then $\lim_{n \to\infty} a_n = \lim_{n \to\infty} b_n$
This was my attempt:
Denote $\lim_{n \to\infty} a_n = l$ and $\lim_{n \to\infty} b_n = m$. Suppose $m \neq n$. Suppose $\epsilon = \frac{l-m}{2}$. By the convergence of $\{a_n\}$ and $\{b_n\}$, and using the specified value of epsilon, for sufficiently large $n$ we have that $\frac{l+m}{2} < a_n < \frac{3l-m}{2}$, and $\frac{3m-l}{2} < b_n < \frac{m+l}{2}$. From this we have
$$0<a_n - b_n < 4\bigg(\frac{l-m}{2}\bigg)$$ $$\rightarrow 0 < a_n - b_n < 4\epsilon$$
But by the density of $\mathbb{R}$, there exists some $r \in \mathbb{R}$ such that $a_n - b_n > r$ for sufficiently large $n$. But this contradicts the fact that $\{a_n - b_n\}$ is a null sequence, hence $l=m$ $$\tag*{$\blacksquare$}$$
I am interested in seeing if there is a proof (and hopefully also verification that mine is correct!) that does not rely on deducing a contradiction from assuming $l \neq m$. This frustratingly seems like one of those 'obvious' statements that when I write out in first order logic I struggle to prove. In particular I couldn't figure out a way to do it directly.
Proof by contradiction really is the most natural approach here. The intuition is simple: if the sequences have different limits, they eventually have to be close to those limits and therefore cannot be close to each other.
It can be done a little more easily, though. Let $\epsilon=\frac13|\ell-m|$. There is an $n_0\in\Bbb N$ such that $|a_n-\ell|<\epsilon$ and $|b_n-m|<\epsilon$ whenever $n\ge n_0$. But then
$$|\ell-m|\le|\ell-a_n|+|a_n-b_n|+|b_n-m|<|a_n-b_n|+2\epsilon\,,$$
for all $n\ge n_0$, so
$$|a_n-b_n|>|\ell-m|-2\epsilon=\epsilon$$
for all $n\ge n_0$, contradicting the assumption that $\langle a_n-b_n:n\in\Bbb N\rangle$ is a null sequence.
Your argument does have some problems. First, you seem to be assuming that $\ell>m$; there’s no real loss of generality if you make this assumption, but you do at least need to say that you’re making it. You’re also apparently assuming at the end that $a_n-b_n$ is positive, which need not be the case. Finally, and most important, you have not actually given any justification for the assertion that there is a real $r$ such that $a_n-b_n>r$ for sufficiently large $n$: this actually is true for $|a_n-b_n|$ and some positive $r$, but this has nothing to do with the density of $\Bbb R$.