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Apologies for my title. I couldn't get all the info in without running out of room. Feel free to edit. I'm just looking for verification of my proof. This problem is from some study materials I am working through, and my solution feels a little too easy. I will state the problem verbatim below, and also provide an answer for critiquing. If it has flaws, I will delete my answer. Others can feel free to answer either way.

Problem: Let $\{x_n\}$ be a sequence in a complete metric space $(X,d)$ and for every $\varepsilon>0$ there exists a convergent sequence $\{y_n\}$ such that $\sup_nd(x_n,y_n)<\varepsilon$. Prove that then $\{x_n\}$ converges.

Here is the answer I came up with:

Let $\varepsilon>0$ be given. Chose a sequence $\{y_n\}$ such that $y_n\to y$ for some $y\in X$ and $\sup_{n\in\mathbb{N}}d(x_n,y_n)<\frac{\varepsilon}{2}$. Such a sequence exists by hypothesis. Since $y_n\to y$, there exists $N\in\mathbb{N}$ such that $\forall n\geq N$, $\;d(y_n,y)<\frac{\varepsilon}{2}$. Then, $\forall n\geq N$, we have by the triangle inequality that:

$$d(x_n,y)\leq d(x_n,y_n)+d(y_n,y)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,$$

and since $X$ is complete, we have that $x_n\to y$, and $\{x_n\}$ is convergent in $(X,d)$.

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    For the proof-verification, you probably should have put your "proof to be checked" in the question, not as answer. – 2017-01-18
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    @ClementC. Oh, good point. Noted. – 2017-01-18
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    Can the downvoter explain? – 2017-01-26
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    I'm afraid most of them don't care enough to (in this case, I really don't see a reason to downvote). – 2017-01-26
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    @ClementC. I know who it was. They downvote me once every few days because I have called them out for never accepting answers, sloppy editing, and a general poor attitude. I'm just getting the vibe out there. – 2017-01-26
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    In that case, you should bring it do a moderator. This sort of behavior is explicitly forbidden by the site's policies. – 2017-01-26
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    I have, but they refuse to do anything. They are sympathetic, and they acknowledge his poor behavior, but apparently there isn't much room for doing anything. It would all be subjective anyway, and I prefer not to bother them. – 2017-01-26
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    In any case, thank you very much for your attentiveness, @ClementC. – 2017-01-26

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The "bug."

The issue with your proof is that $y$ depends on $\varepsilon$ itself, so the end statement you show has the quantifiers wrong: you show "$\forall \varepsilon > 0,\ \exists y,\ [\cdots]$" instead of "$\exists y,\ \forall \varepsilon > 0,\ [\cdots]$."

One thing that should be alarming is that nowhere have you actually used the completeness of $X$. You wrote "since $X$ is complete," but you are not using it to argue anything. But one rarely throw in unnecessary assumptions for the sake of it.

A proof.

Let $\varepsilon > 0$ be an arbitrary positive number. Choose a convergent sequence $(y^{(\varepsilon)}_n)_{n\geq 0}$ such that $$\sup_{n\geq 0} d(x_n,y^{(\varepsilon)}_n) \leq \frac{\varepsilon}{3} \tag{1} $$

Since $(y^{(\varepsilon)}_n)_{n\geq 0}$ is convergent, it is Cauchy: let $n_\varepsilon\geq 0$ be such that for all $n\geq n_\varepsilon$ and $m\geq 0$, $$d(y^{(\varepsilon)}_{n+m},y^{(\varepsilon)}_n) \leq \frac{\varepsilon}{3}\tag{2}$$

Now, for any $n\geq n_\varepsilon$ and $m\geq 0$, using (1), (2), and the triangle inequality $$ d(x_{n+m},x_n) \leq d(x_{n+m},y^{(\varepsilon)}_{n+m}) + d(y^{(\varepsilon)}_{n+m},y^{(\varepsilon)}_n) + d(y^{(\varepsilon)}_n,x_n) \leq \frac{\varepsilon}{3} +\frac{\varepsilon}{3} +\frac{\varepsilon}{3} = \varepsilon $$ showing that $(x_n)_{n\geq 0}$ is Cauchy. By completeness of $X$, it it thus convergent.

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    Thanks very much, Clement. I'll go through this in detail in a little while and make sure it makes sense to me. – 2017-01-18
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    Having read it, so my mistake was trying to relate the sequence $\{x_n\}$ to the point that $\{y_n\}$ converges to? It looks you did what I did but just used the Cauchy-ness of the $y_n$ sequence instead. Am I understanding correctly? – 2017-01-18
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    Yes, exactly. ${}$ – 2017-01-18