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Let me just give the proof I found from a book, which seems incomplete:

First, for a particular value of $r\in\mathbb R$, $f^{-1}(]r-\frac{1} {k},r+\frac{1}{k}[), k\in\mathbb Z_{++}$ is an open subset of $\mathbb R^\omega$, which, under the product topology, is of the form $\mathcal O_k\times\mathbb R^\omega$, where $\mathcal O_k$ is an open subset of $\mathbb R^{n_k}$ for some $n_k\in\mathbb Z_{++}$.

Now, for any $k'>k$, we have $f^{-1}(]r-\frac{1}{k'},r+\frac{1}{k'}[)\subset f^{-1}(]r-\frac{1}{k},r+\frac{1}{k}[)$, and therefore "it is easy to deduce that" for any $f(x,y)=r$, we have $x\in\mathcal O_k$ and $y$ is any point in $\mathbb R^\omega$, and therefore, $f(x,y)=f(x)$ locally.

My problem with the last deduction is that: as we shrink the interval centered at $r\in\mathbb R$, the preimage of the interval may involve different (though finite) number of coordinates, for example, an increasing number of coordinates. It is not so obvious this remains finite when we take the limit $k\to\infty$. I have checked Wilansky and Munkres and didn't find an answer (Munkres is more obsessed with maps into products).

Any suggestion or references are welcome.

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You are right. The given argument is wrong, when we shrink the neighbourhood of $r$, the set of components constrained by $f^{-1}\bigl(\bigl]r-\frac{1}{k}, r + \frac{1}{k}\bigr[\bigr)$ grows, and in the limit often exhausts all of $\omega$.

I don't know what "$f(x,y) = f(x)$ locally" is supposed to mean, so I can't tell if the assertion is true for the intended meaning. But of course a continuous $f \colon \mathbb{R}^{\omega} \to \mathbb{R}$ can depend on all coordinates in such a manner that all sections keeping all but one coordinate constant are injective, e.g.

$$f(x) = \sum_{n = 0}^{\infty} 2^{-n} \frac{x_n}{1 + \lvert x_n\rvert}.$$

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    Thanks for the reply. But then I am very confused by your second sentence. Anyway, I think you are right and the author must have referred to a completely different topology. I consider it product topology because the author described its basis as those for a product topology. Maybe it is not. – 2017-01-11
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    Roughly, what I'm saying is that - using the notation of the quote in your question - the sequence $(n_k)$ (which gives the number of coordinates constrained in $\mathcal{O}_k \times \mathbb{R}^{\omega}$) typically is unbounded, $\lim\limits_{k\to\infty} n_k = +\infty$. – 2017-01-11
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    Yes. I get that part. I think now I get the rest. Thanks. Let me just accept your answer now. In fact, just now, I found a paper written by the author about the same subject. He said Fréchet topology...I was shocked how he omitted that word in his entire book... – 2017-01-11
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    What Fréchet topology is the author referring to? The product topology makes $\mathbb{R}^{\omega}$ a Fréchet space (in the sense of functional analysis), so they might just mean the product topology. Also, $T_1$-spaces are sometimes called Fréchet spaces, and there might be a technical meaning of Fréchet topology that I've not yet encountered. – 2017-01-11
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    I was just checking that as you made the comment. Again I think you are right, but the whole thing just doesn't fit. He referred to GTM14. I am still struggling to understand what he meant (I'm only an engineer). And by the way, I am lucky to meet you: I realize that you used to work on topological vector space :-). – 2017-01-11