Here is Prob. 8, Sec. 20, in the book Topology by James R. Munkres, 2nd edition:
Let $X$ be the subset of $\mathbb{R}^\omega$ consisting of all sequences $x$ such that $\sum x_i^2$ converges. Then the formula $$ d( \mathbf{x}, \mathbf{y} ) = \left[ \sum_{i=1}^\infty \left( x_i - y_i \right)^2 \right]^{1/2} $$ defines a metric on $X$. . . . On $X$ we have the three topologies it inherits from the box, uniform, and product topologies on $\mathbb{R}^\omega$. We have also the topology given by the metric $d$, which we call the $\ell^2$-topology. . . .
(a) Show that on $X$, we have the inclusions $$ \mbox{ box topology } \supset \ell^2\mbox{-topology } \supset \mbox{ uniform topology}. \tag{0} $$
This I've managed to do.
(b) The set $\mathbb{R}^\infty$ of all sequences that are eventually zero is contained in $X$. Show that the four topologies that $\mathbb{R}^\infty$ inherits as a subspace of $X$ are all distinct.
My Attempt:
The set $\prod_{k=1}^\infty \left( -1/k^3, 1/k^3 \right)$, for example, is open in the box topology on $\mathbb{R}^\omega$; so the set $\mathbb{R}^\infty \cap \left[ \prod_{k=1}^\infty \left( -1/k^3 , 1/k^3 \right) \right]$ is open in the corresponding subspace topology on $\mathbb{R}^\infty$; the zero sequence $\mathbb{0} \colon= (0, 0, 0, \ldots)$ is in this set, but for any $\delta > 0$, we note that if $n$ is any natural number such that $n > 2/\delta$, then the element $$\mathbf{x} \colon= \left(\underbrace{ 1/n^3, \ldots, 1/n^3 }_{n \mbox{ terms } }, 0, 0, 0, \ldots \right)$$ is in $$ \mathbb{R}^\infty \cap B ( \mathbf{0}, \delta) \colon= \left\{ \ \mathbf{p} \in \mathbb{R}^\infty \ \colon \ d( \mathbf{p}, \mathbf{0} ) < \delta \ \right\}, $$ because $$ d ( \mathbf{x}, \mathbf{0} ) = \sqrt{ \sum_{i=1}^n 1/n^3 } = 1/n < \delta/2 < \delta, $$ but $\mathbf{x} \not\in \prod_{k=1}^\infty \left( -1/k^3, 1/k^3 \right),$ because $ 1/n^3 \not\in \left( -1/n^3, 1/n^3 \right)$, and hence $\mathbf{x} \not\in \mathbb{R}^\infty \cap \left[ \prod_{k=1}^\infty \left( -1/k^3, 1/K^3 \right) \right]$.
Thus although $\mathbf{0} \in \mathbb{R}^\infty \cap \left[ \prod_{k=1}^\infty \left( -1/k^3, 1/K^3 \right) \right]$, no $d$-metric open ball in $\mathbb{R}^\infty$ about $\mathbf{0}$ is contained in this set, showing that this set, which is open in the "box topology" on $\mathbb{R}^\infty$ is not open in the "$\ell^2$-topology".
From part (a), we can conclude that the corresponding subspace topologies on $\mathbb{R}^\infty$ also satisfy these very inclusion relations as in (0) above. So we can conclude that on $\mathbb{R}^\infty$, the "box topology" is strictly finer than the "$\ell^2$-topology".
Is what I have done so far correct?
The set $$ \mathbb{R}^\infty \cap B ( \mathbf{0}, 1 ) = \left\{ \ \mathbf{x} \in \mathbb{R}^\infty \ \colon \ d( \mathbf{x}, \mathbf{0} ) < 1 \ \right\}$$ is open in the "$\ell^2$-topology" on $\mathbb{R}^\infty$ and contains the point $\mathbf{0}$. However, for any real number $\delta > 0$, if we take a natural number $n > 4/\delta^2$, then the point $$ \mathbf{p} \colon= \left( \underbrace{ \delta/2, \ldots, \delta/2}_{ n \mbox{ terms } }, 0, 0, 0, \ldots \right) $$ is in $$ \mathbb{R}^\infty \cap B_{ \bar{\rho} } ( \mathbf{0}, \delta ) \colon= \left\{ \ \mathbf{x} \in \mathbb{R}^\infty \ \colon \ \bar{\rho} ( \mathbf{x}, \mathbf{0} ) < \delta \ \right\}, $$ because $$ \bar{\rho} ( \mathbf{p}, \mathbf{0} ) \leq \delta/2 < \delta, $$ but $$ d( \mathbf{p}, \mathbf{0} ) = \frac{\delta}{2} \sqrt{ n} > 1,$$ and so $$ \mathbf{p} \not\in \mathbb{R}^\infty \cap B ( \mathbf{0}, 1 ).$$
Thus, although the point $\mathbf{0}$ is in the set $\mathbb{R}^\infty \cap B ( \mathbf{0}, 1 )$, no $\bar{\rho}$-metric open ball in $\mathbb{R}^\infty$ about $\mathbf{0}$ is contained in this set, showing that this set, which is open in the "$\ell^2$-topology" on $\mathbb{R}^\infty$, is not open in the "$\bar{\rho}$-topology"; therefore the former topology is strictly finer than the latter one.
Is this portion of my proof correct?
The product topology on $\mathbb{R}^\omega$ is the same as the topology induced by the metric $D$ on $\mathbb{R}^\omega$ given by the formula $$ D ( \mathbf{x}, \mathbf{y} ) = \sup \left\{ \ \frac{ \min \left\{ \ \left\lvert x_n -y_n \right\rvert, \ 1 \ \right\} }{ n } \ \colon \ n \in \mathbb{N} \ \right\}. $$
Now the set $$\mathbb{R}^\infty \cap B_{\bar{\rho}} ( \mathbf{0}, 1/2 ) \colon= \left\{ \ \mathbf{x} \in \mathbb{R}^\infty \ \colon \ \bar{\rho} ( \mathbf{x}, \mathbf{0} ) < 1/2 \ \right\}$$ is open in the "uniform topology" on $\mathbb{R}^\infty$ and contains the point $\mathbf{0}$. However, for any real number $\delta > 0$, if we take $n$ to be any natural number such that $n > 1/\delta$, then the point $$ \mathbf{p} \colon= \left( \underbrace{ 0, \ldots, 0}_{ n-1 \mbox{ terms } }, 1, 0, 0, \ldots \right)$$ is in the set $$\mathbb{R}^\infty \cap B_D ( \mathbf{0}, \delta) \colon= \left\{ \ \mathbf{x} \in \mathbb{R}^\infty \ \colon \ D( \mathbf{x}, \mathbf{0} ) < \delta \ \right\}, $$ because $$ D( \mathbf{p}, \mathbf{0} ) = \frac{1}{n} < \delta, $$ but $$ \bar{\rho}( \mathbf{p}, \mathbf{0} ) = 1 > \frac{1}{2} $$ so that $$ \mathbf{p} \not\in \mathbb{R}^\infty \cap B_{\bar{\rho}} (\mathbf{0}, 1/2 ),$$ thus showing that, although the point $\mathbf{0}$ is in the set $\mathbb{R}^\infty \cap B_{\bar{\rho}} (\mathbf{0} , 1/2 )$, for no $\delta > 0$ is the set $\mathbb{R}^\infty \cap B_D ( \mathbf{0}, \delta)$ contained in $\mathbb{R}^\infty \cap B_{\bar{\rho}} (\mathbf{0} , 1/2 )$.
Thus the set $\mathbb{R}^\infty \cap B_{\bar{\rho}} (\mathbf{0} , 1/2 )$, which is open in the "unifrom topology" on $\mathbb{R}^\infty$, is not open in the "product topology" on $\mathbb{R}^\infty$.
Since the uniform topology on $\mathbb{R}^\omega$ is (strictly) finer than the product topology (i.e. the $D$-metric topology) on $\mathbb{R}^\omega$, so the subspace topology on $\mathbb{R}^\infty$ corresponding to the former topology is finer than the subspace topology corresponding to the latter, and, on the basis of what we have shown in the last few paragraphs, the "unifrom topology" on $\mathbb{R}^\infty$ is strictly finer than the "product topology".
Is this part of my proof correct?
Therefore, we have shown that all the four topologies in question are distinct.
Is my proof accurate? If so, is my presentation clear enough too?