Is this alternative proof of Theorem 3.7 ("Baby" Rudin, Ch. 3) correct and, if so, well written? Rudin, in his Principles of Mathematical Analysis, proves the following theorem:


  The subsequential limits of a sequence $\{p_n\}$ in a metric space $X$ form a closed subset of $X$.


I've tried to come up with a different proof, but I have a few doubts about its correctness:

Let $E$ be the set of the subsequential limits of $\{p_n\}$ and $q$ a limit point of $E$. We have to prove that $q\in E$.

Since $q$ is a limit point of $E$, there exists a sequence $\{a_n\}$ in $E$ that converges to $q$ (this is a theorem that Rudin proves earlier in the book). Let $V_n$ be the open ball of center $a_n$ and radius $d(q,a_n)$.

Since there is a subsequence of $\{p_n\}$ which converges to $a_n$, $V_n \cap p(\mathbb{N})$, where $p(\mathbb{N})$ is the range of the sequence $\{p_n\}$, is nonempty for every $n\in \mathbb{N}$; that is, $V_n$ always contains at least one term of the sequence. 

Therefore, by the axiom of countable choiche, there exists a sequence $\{q_n\}$ such that $q_n \in V_n\cap p(\mathbb{N})$ for every $n$. This happens to be a subsequence of $\{p_n\}$.

Now it is left to prove that $q_n$ converges to $q$: let $\epsilon &gt;0$ be given; then there is $N$ such that $d(q,a_n)&lt;\epsilon$ if $n&gt;N$. We also find that $d(a_n,q_n)&lt;d(q,a_n)$; thus, by the triangle inequality, $d(q,q_n)\leq d(a_n,q_n)+d(q,a_n)&lt;2\epsilon$. Hence the thesis.

I realize that it's definitely more complicated that the one offered by the author, but is it sound nonetheless? Is it written in a fashion that would be accepted in an exam?
Here is how to prove it without the axiom of choice:...[0]:.. Let $q$ be a limit point of $E$...[1]:Observe that for any member of $E$ there is a strictly increasing $f : N \to N$ such that  $(p_{f(n)})_{n \in N}$ converges to it...[2]:.. For all $r&gt;0$, the set  $S(r)= \{ n \in N : p_n \in B(q,r) \}$ is infinite because...[2a]:.. B(q,r) contains some $q^* \in E$ and,by [1], $q^*$is the limit of a sequence $(p_{f(n) : n \in N})$ for some strictly increasing $f:N \to N$, and ...[2b]:.. There exists, by the Triangle Inequality, some $s&gt;0$ for which $B(q^*,s) \subset B(q,r)$, so... [2c}:..The infinite set $\{ n \in N : p_{f(n)} \in C \}$ is a subset of $S(r)$....[3]:..Now define $g:N \to N $ recursively by letting $g(0)= \min \{ k: p_k \in B(q,1) \}$ and $g(n+1)= \min \{ k&gt;g(n) : p_k \in B(q,2^{-n-1} \}$. Then $g$ exists by [2] and is strictly increasing and $(p_{g(n)})_{n \in N}$ converges to $q$.
Here is how to prove it without the axiom of choice:...[0]:.. Let $q$ be a limit point of $E$...[1]:Observe that for any member of $E$ there is a strictly increasing $f : N \to N$ such that  $(p_{f(n)})_{n \in N}$ converges to it...[2]:.. For all $r&gt;0$, the set  $S(r)= \{ n \in N : p_n \in B(q,r) \}$ is infinite because...[2a]:.. B(q,r) contains some $q^* \in E$ and,by [1], $q^*$is the limit of a sequence $(p_{f(n) : n \in N})$ for some strictly increasing $f:N \to N$, and ...[2b]:.. There exists, by the Triangle Inequality, some $s&gt;0$ for which $B(q^*,s) \subset B(q,r)$, so... [2c}:..The infinite set $\{ n \in N : p_{f(n)} \in C \}$ is a subset of $S(r)$....[3]:..Now define $g:N \to N $ recursively by letting $g(0)= \min \{ k: p_k \in B(q,1) \}$ and $g(n+1)= \min \{ k&gt;g(n) : p_k \in B(q,2^{-n-1} \}$. Then $g$ exists by [2] and is strictly increasing and $(p_{g(n)})_{n \in N}$ converges to $q$.