Suppose that $f: \mathbb R\to \mathbb R^\omega$, where $\omega$ is countably infinite index set, is defined by $f(t)= (t,2t,3t,...)$.
The question is to discuss continuity of $f$ in the following three cases: $1) \mathbb R^\omega$ is with product topology, $2) \mathbb R^\omega$ is with the uniform topology, $3)\mathbb R^\omega$ is with the box topology.
Case $1:$ In this case, $f$ is continuous since each of its component functions are continuous.
Case $2:$ This topology is defined as the topology induced by the metric $\rho:\mathbb R^\omega\times \mathbb R^\omega\to [0,\infty), \rho(x,y)=\sup\{\min(1, |x_i-y_i|): i\in \mathbb N\}$.
Let $B_\rho(x, r), r<1$ be a basis element of the uniform topology. We wish to show that $f^{-1}(B_\rho(x, r))\subset \mathbb R$ is open. It is trivial if $f^{-1}(B_\rho(x, r))$ is empty or all of $\mathbb R$ so suppose this to be non empty and properly contained in $\mathbb R$. Let $y\in f^{-1}( B_\rho(x, r)).$ It is to be shown that there exists a $\delta>0$ such that $(y-\delta, y+\delta)\subset f^{-1}(B_\rho(x, r)).$
We have $\rho (f(y),x))\lt r\implies |\pi_i(f(y))-x_i|\lt r\,\, \forall i\implies |iy-x_i|\lt r\implies |y-\frac {x_i}i|\lt \frac r{i}.$ It is tempting to assume $\delta=\min_i \frac{x_i}i$ but the problem is that such a $\delta$ may not be well defined. This makes me think that $f$ is not continuous. Nonetheless, I'm stuck here.
Case $3:$ Consider the open set $\Pi U_i$ in the co-domain, where $U_i:=(-\frac 1i, \frac 1i)$. It is clear that $f^{-1}(\Pi U_i)\ni 0$ so if $f$ were continuous then $f^{-1}(\Pi U_i)$ should have been open in $\mathbb R$,i.e, there must exist a $\delta>0$ such that $(-\delta, \delta)\subset f^{-1}(\Pi U_i)\implies f((-\delta, \delta))\subset \Pi U_i\implies (-i\delta, i\delta)\subset(-\frac 1i, \frac 1i)\, \forall i\implies \delta\lt \frac 1i\, \forall i$, which is not possible (as it would violate Archimedean property). So $f$ is not continuous here.
It is known that product topology is contained in the uniform topology which in turn is contained in the box topology.
If $f$ were not continuous in the uniform topology then it will not be continuous in the box topology either. If $f$ were continuous in the uniform topology then nothing can be said about continuity of $f$ in the box topology without further investigation. The point is that: continuity of $f$ in case $2$ does not seem to be deducible directly from case $1$ and case $3$.
So how do we conclude continuity (or non continuity) of $f$ is case $2$?
With help from @MartinSleziak, here is a solution:
It is clear that $f(0)=0$. Let's check the behavior of $f$ around $0$. For any $x\ne 0$,
$\rho (f(x), f(0))=\sup \{\min (1, |\pi_i(f(x))-\pi_i(f(0))|): i\in \mathbb N\}=\sup \{\min (1, |ix|):i\in \mathbb N\}=1.$
So $f$ is not continuous at $0$.