Is there a clever way to avoid choice in Riesz Representation Theorem?

Rudin RCA p.43

Riesz Representation for LCH:

Let $X$ be a locally compact Hausdorff space, and let $\Lambda$ be a positive linear functional on $C_c(X)$. Then there exists a $\sigma$-algebra $\mathfrak{M}$ on $X$ which contains all Borel sets in $X$, and there exists a unique positive measure $\mu$ on $\mathfrak{M}$ which represents $\Lambda$ in the sense that $\Lambda f = \int_X f d\mu$ for every $f\in C_c(X)$.

Since Urysohn's lemma can be proven without choice when $X$ is locally compact Hausdorff and second countable, i thought Riesz-Representation theorem could be proven without choice if $X$ is, additionally, second countable.

All the other parts of the Rudin's proof works fine without choice, but there is one part i'm stuck.

Below is Rudin's argument:

(The notation $f\prec V$ will mean that $V$ is open, that $f\in C_c(X)$, $0≦f≦1$, and that $\text{supp}f \subset V$)

For every open set $V$ in $X$, define $\mu(V)=\sup\{\Lambda f : f\prec V\}$.

Then define $\mu(E)=\inf\{\mu(V):E\subset V, V \text{ open}\}$ for every subset $E$ of $X$. (It is consistent with the above definition. Then it can be shown that "$E\subset F \Rightarrow \mu(E)≦\mu(F)$"

Now, let $\{E_n\}$ be a sequence of subsets of $X$, and I need to prove "$\mu(\bigcup_{n\in\mathbb{N}} E_n)≦\sum_{n=1}^{\infty} \mu(E_n)$".

Since Urysohn's Lemma holds for locally compact second countable space, it can be proven that $\mu(V_1\cup V_2)≦\mu(V_1)+\mu(V_2)$ for every open sets $V_1,V_2$.

Suppose $\mu(E_n)<\infty$ for all $n\in\mathbb{N}$.

Fix $\epsilon>0$.

Then, by definition, there exist open sets $V_n$ such that $E_n\subset V_n$ and $\mu(V_n)<\mu(E_n) + 2^{-n}\epsilon$ (**Axiom of Countable Choice is used here)

Put $f\prec \bigcup_{n\in\mathbb{N}} V_n$. Since $\text{supp} f$ is compact there exists a finite subcover $\{V_1,...,V_m\}$ of $\text{supp} f$, hence $\Lambda f≦\mu(\bigcup_{i=1}^m V_i)≦\sum_{i=1}^m \mu(V_i)≦\sum_{i=1}^{\infty} \mu(E_i) + \epsilon$.

Thus $\mu(\bigcup_{n\in\mathbb{N}} V_n)≦\sum_{n=1}^{\infty} \mu(E_n)+\epsilon$. Since $\bigcup_{n\in\mathbb{N}} E_n \subset \bigcup_{n\in\mathbb{N}} V_n$, the proof is done.

I struggling to avoid that 'countable choice'.. Is there a clever way to avoid this choice?($X$ is assumed to be locally compact and second countable)

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Without the Axiom of Countable Choice, one cannot prove that the Lebesgue measure on $\mathbb{R}^{n}$ is $\sigma$-additive. Hence, one runs into such undesirable circumstances when trying to avoid the Axiom of Choice altogether. –  Haskell Curry Jan 20 '13 at 9:02

In particular you might be interested in theorem $\mathbf{564I}$.
@commenter: No, the other way around. Without choice things are more complicated; not that with choice they are all that easy, but at least the basic levels can be dealt with quite smoothly (e.g. completion of a $\sigma$-additive measure is $\sigma$-additive). –  Asaf Karagila Jan 21 '13 at 22:59