Proving that a sequence of reals, $(x_n)$, converges to $x$ iff $\lim_{n \to \infty} d(x_n, x) = 0$ In this situation, $d(x, y) = |x - y|$, the standard metric.
I think this is a very basic proof, but I want to make sure I am being thorough enough.
Forward direction: $(x_n)$ converging to x means, by definition, that for any $\epsilon > 0$ there exists $N > 0$ such that whenever $n \ge N$, $|x_n - x| \lt \epsilon$. This result can also be written as $d(x_n, x) \lt \epsilon$, or $d(x_n, x) - 0 \lt \epsilon$. Therefore $\lim_{n \to \infty} d(x_n, x) = 0$.
Backward direction: $\lim_{n \to \infty} d(x_n, x) = 0$ means that for any $\epsilon > 0$ there exists $N > 0$ such that $n \ge N$ implies $d(x_n, x) - 0 = d(x_n, x) \lt \epsilon$. So, $|x_n - x| \lt \epsilon$. So, $(x_n)$ converges to $x$.
Is it this simple? Or am I missing any steps? Thank you so much for any help. I am still relatively new to mathematical proof writing.
Note - this is Lemma 1.1.1 (and exercise 1.1.1) in Tao's Analysis 2, and I just want to make sure I am being precise enough in the answer.

Your proof is fine. However you need to make sure to put absolute when the limit is not zero as;
$$|d(x_n,x)- L| < \epsilon.$$
When $L=0$, (as in your case), the absolute value dose not matter.
Notation:
Your theorem is (clear fact) when the metic is the usual matrix, namely
$$d(x, y) = |x-y|.$$
But in the case of $d(x,y)$ is a general metric, it is very important fact which assures that whenever we have a convergent sequence in the metric space (as instance the space of continuous functions on closed interval with a norm, $(C[a,b], d=\| f-g\|_\infty)$, the theorem proves that the distance between the sequence and its limit according to $d$, namely $d(x_n,x)$ is convergent to zero. This theorem is very important in proving many results. See the proof of completeness of the Euclidian space $\mathbb{R}^n$ and $\mathbb{C}^n$ as example.
