This is Munkres §44 Problem 3. (a), which asks me to prove that if $\mathbb{R}^ω$ is given the product topology, then there is no continuous surjective map $f: \mathbb{R} → \mathbb{R}^ω$. The hint asks to me show that $\mathbb{R}^ω$ is not $\sigma$-compact, since the continuous image of a $\sigma$-compact space is $\sigma$-compact.
My attempt: I tried to find some property which $\sigma$-compactness implies, but which $\mathbb{R}^ω$ doesn't possess. $\sigma$-compactness clearly implies Lindelöf, but $\mathbb{R}^ω$ is Lindelöf, being metrizable and separable. Then I thought of local compactness. $\mathbb{R}^ω$ is not locally compact, being infinite-dimensional. And I found this theorem, which says that a Hausdorff, Baire space that is also $\sigma$-compact, must be locally compact at at least one point. Since $\mathbb{R}^ω$ is a Baire space (I think so, it is completely metrizable), it solves my problem, but I'm not supposed to know about Baire spaces (they appear later in Munkres' book). How can I get around this? I think Munkres' intention is for me to directly prove that $\mathbb{R}^ω$ is not $\sigma$-compact.
Since the projection map onto each coordinate is continuous, it must have compact image when restricted to a compact subset of $\mathbb{R}^\omega$, so for any compact $K \subseteq \mathbb{R}^\omega$, there must exists a sequence $R_i > 0$ s.t. $K \subseteq \prod_i [-R_i, R_i]$. So if $\mathbb{R}^\omega$ is $\sigma$-compact, then you can choose an increasing sequence of compact subsets $K^j$ whose union is the entire space. Each $K^j$ then is a subset of $\prod_i [-R^j_i, R^j_i]$, with $R^j_i$ increasing as $j$ increases for each fixed $i$. You can then explicitly construct an element in $\mathbb{R}^\omega$ that is not contained in any $K_j$. Indeed, you can define such an $(r_i) \in \mathbb{R}^\omega$ by requiring $r_i \notin [-R^i_i, R^i_i]$ for all $i$. But this contradicts with the assumption that the union of $K_j$ is the entire space, so we must have $\mathbb{R}^\omega$ is not $\sigma$-compact.