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Let $(M,d)$ be a metric space and $A\subset M$ be a discrete set. I want to show that if $A$ is compact, then $A$ is finite using sequences. I proceeded as follows, but I'm unsure about one step of my proof:

Suppose, for the sake of contradiction, that $A$ is infinite. In that case, one can form one sequence $(x_n)$ with distinct points of $A$. Since $A$ is compact, $(x_n)$ has a subsequence $(x_{n_k})$ which converges to a point $a\in A$. In that case, given $\epsilon > 0$ there is $n_0 \in \mathbb{N}$ such that $n_k > n_0$ implies $x_{n_k} \in B_\epsilon(a)$ the open ball centered at $a$ with radius $\epsilon$. Since all the points of the sequence are distinct, this means that there are infinite points of $A$ inside that open set containing $a$ which are different than $a$. It follows that $a$ is not an isolated point, which is a contradiction since $A$ is discrete, and it follows that $A$ is finite.

Now, what I'm unsure is that I said "since it is infinite we can construct one sequence with all elements different than each other". Although intuitively I felt this was necessary, I can't find any way to see why we would need all elements of the sequence different than each other. In truth, I would need only to show at least one element different than $a$ exists inside the neighborhood in order to produce the contradiction.

With this, is my proof correct? If so, how can I justify the need to have a sequence with all elements distinct from one another?

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Yes, you can try to justify it with Axiom of Choice.

Edit: I'm not sure if I understand you correctly. You'd need 1 element different from $a$ in EVERY nbh. of $a$ so you'd need a sequence with infinite distinct elements anyway.

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