Is this proof correct?
If $\{f_{n}\}$ is a sequence of functions in $C(X,Y)$, $X$ compact, $Y$ complete, and the sequence converges, to $f$, then $K=(\bigcup\{f_n\})\cup \{f\}$ is closed.
Proof.
Assume that a sequence $f_{n_i}$ is in $K$ and it converges.
Either this sequence consist of a finite amount of elements and some of them repeats themselves, or there are infinitely many different elements in the sequence.
In the first case, in order to get convergence, we must obviously have a constant sequence after some time, and it converges to an element in $K$.
In the second case $\{f_{n_i}\}$ consists of infinitely many different elements, then we can create a subsequence where we pick out elements such that $n_i>n_j$ if $i>j$, we are able to do this because $\{n_i\}$ has infinite many different elements, so if we have $n_{i_1},n_{i_2},...n_{i_N}$ there is always possible to find and $n_{i_{N+1}}$ which is bigger than all the others and we also have that $i_{N+1}$ is bigger than all the other i's. But this subsequence is a subsequence of ${f_n}$ and it converges to $f$. Hence $K$ is closed.
I am a little unsure in constructing the subsequence, and if what I did was legal. I mean could I do it the way I did, and do you have a simpler way to write how you would create the subsequence?
UPDATE Maybe the proof above was a little weird. i tried another easier one. I want to prove that a sequence in K can not converge to any point outside K. Instead of showing that the limit is inside K, I use contradiction to show that the limit is not outside K.
Proof 2
Assume for contradiction that $f_{n_i}$ converges to $g$ and g is not in K. Create a subsequence of $f_{n_i}$, since there is a distance $\rho(f_1,g)$, choose $n_{i_1}$ so far out in the sequence of $\{f_{n_i}\}$ such that the remaining terms in the sequence have distance less that $\rho(f_1,g)$. Now only consider the part of the sequence $\{f_{n_i}\}$ after $f_{n_{i_1}}$. Then we repeat the argument for $f_2$, and we get an element $f_{n_{i_2}}$. We now have a subsequence $\{f_{n_{i_k}}\}$ of $\{f_{n_i}\}$. This subsequence must conerge to $g$. However this subsequence must also converge to f, because given an $\epsilon$ there is an N for the sequence $\{f_n\}$, where if we look at terms after N, we are as close as $\epsilon$ to f for the remaining terms, but this means that the remaining terms after $f_{n_{i_N}}$ must also be this close to f. Hence we have our contradiction.
Is this one correct?
PS: It may be that the whole statement is false, but I doubt that very much.
The proofs make sense; I think I like the first one more. As commenters pointed out, this result is true in every metric space. In fact, it is true in every Hausdorff topological space.
The proof could be made simpler by focusing on the complement of the set $K$, to show it's open. Indeed, take $g$ not in $K$. Since $f\ne g$, there are disjoint neighborhoods $U,V$ of $f$ and $g$ respectively. Since $f_n\to f$, all but finitely many elements of the sequence lie in $U$. Hence, only finitely many lie in $V$. Shrink the neighborhood further to leave them out (in a metric space, you'd use the minimum of distances for this; in a Hausdorff topological space, finite intersection of neighborhoods would be involved). Conclusion: $g$ has a neighborhood that is disjoint from $K$.
As a general fact, proofs that focus on openness rather than closedness tend to involve less manipulation of subsequences.