A net $\varphi : [0, \omega_1) \to M$ on a metric space $M$ converges $\iff \varphi$ is eventually constant I want to prove that 

If $M$ is a metric space, then a net $\varphi : [0, \omega_1) \to M$ converges if and only if $\varphi$ is eventually constant. ($[0, \omega_1)$ is the set of ordinals less than $\omega_1$, where $\omega_1$ is the first uncountable ordinal, with the order topology).

$(\Leftarrow)$ is clear.
My trouble is with $(\Rightarrow)$. Suppose that $\varphi \rightarrow x$.
Let $U = \{ \alpha \in [0, \omega_1) \, \mid \varphi(\alpha) \neq x \}.$
If $U = \emptyset $ we are done, so assume $U \neq \emptyset$. Since $[0, \omega_1)$ is well ordered, there is a $\alpha_1 \in U$ such that $ \alpha_1 \leq \alpha \, \, \forall \alpha \in U.$
If $ U \setminus \alpha_1 = \emptyset$ we are done. If not, we repeat the procedure.
It is clear that if $card(U) = n$ for some $n \in \omega$ then we only have a finite set of ordinals $\{\alpha_1, \ldots, \alpha_n\}=U$ so $\varphi$ is constant for $\alpha > \alpha_n.$
If $U$ is countably infinite, then $U = \{\alpha_n\}_{n \in \omega}$ and since $[0, \omega_1)$ is sequentially compact and the $\alpha_n$'s are countable ordinals  we obtain a convergent subsequence $\{\alpha_{n_k}\}$ of $U$ to some countable ordinal $\beta \in [0, \omega_1)$ $[\beta = \cup_{n_k \in \omega}\alpha_{n_k}???],$ hence $\varphi$ is constant for all $ \alpha > \beta.$

But what if $U$ is uncountable? 

Idea: If $U = [0, \omega_1),$ then since $\varphi \rightarrow x$, for every $n \in \omega$ there is $\alpha_n \in U$ such that $\alpha \geq \alpha_n \implies \varphi(\alpha) \in B_{\frac{1}{n}}(x).$ This is a countable sequence, so since $[0, \omega_1)$ is sequentially compact, there is a subsequence converging to a countable ordinal $\beta \in [0, \omega_1)$.
Clearly, $\beta$ satisfies that $\phi(\beta) \in B_{\frac{1}{n}}$(x) for every $n \in \omega.$

We can conclude that $\varphi(\beta)= x$ right?

For if $\varphi(\beta) \neq x$, then taking $n$ such that $0 < \frac{1}{n} < d(x, \varphi(\beta)),$ we would have that $\varphi(\beta) \not \in B_{\frac{1}{n}}(x),$ contradicting the hypothesis.
I feel pretty confident about this, but If someone sees an error in the argument, please comment. Thanks.

For each $n$, the set $\left\{\alpha\in[0,\omega_1):d(\varphi(\alpha),x)>1/n\right\}$ is countable, so $\left\{\alpha\in[0,\omega_1):\varphi(\alpha)\neq x)\right\}$ is also countable, so it is not cofinal in $\omega_1$.
