Proof that if every rational limit is a sublimit, then every real number is a sub limit

Question

I need to prove or disprove the following statement:

Let $ a_n $ be a sequence such that for every $ q \in \mathbb{Q} $, $ q $ is a sublimit of $ a_n $. Then every real number is also a sublimit. Is my proof correct?

Let $ c \in \mathbb{R} $ and suppose, for the sake of contradiction, that $ c $ is not a sublimit of $ a_n $.

Therefore, there exists a $ \delta > 0$ such that there is only a finite number of entries from $ a_n $ in $ (c - \delta, c + \delta) $.

$ \mathbb{Q} $ is dense in $ \mathbb{R} $ and therefore there exists a $ q \in \mathbb{Q} $ such that $ c - \delta < q < c $.

Choose $ \delta_0 = distance(q, c- \delta) $. Therefore, there is only a finite number of entries from $ a_n $ in $ (q - \delta_0, q + \delta_0) $ and therefore $ q $ is not a sublimit of $ a_n $ - contradiction.

If $ a_n $ is a sequence and there exists a subsequence $ a_{n_k} $ such that $ a_{n_k} \to L$, then $ L$ is a partial limit of $ a_n $. — talopl, Apr 07 '23 at 13:13
This is correct. More generally, the set of partial limits of a sequence is closed, i.e if $(b_n)$ is a sequence of partial limits of $a_n$, and $b_n\to b$, then $b$ is also a partial limit of $a_n$. — Mark, Apr 07 '23 at 13:21
Yes it’s correct. Using the countability of Q you can actually construct such a sequence. — Vivaan Daga, Apr 07 '23 at 13:23
Regarding @Mark's comment, see Proof that the set of subsequential limits is closed. Conversely, we have Given closed $C \subseteq \mathbb{R}$ find a sequence with subsequences convergent to every point in $C$ and nowhere else AND Any closed subset of $\mathbb C$ is the set of limit points of some sequence. — Dave L. Renfro, Apr 07 '23 at 14:08

Proof that if every rational limit is a sublimit, then every real number is a sub limit

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