Let $f:\mathbb{R}\to \mathbb{R}^{\omega}$ defined by $f(t)=(t,2t,3t,\dots)$ where $\mathbb{R}$ is with standard topology and $\mathbb{R}^{\omega}$ with uniform topology. Is $f$ continuous or not?
Let's recall what uniform topology is? Define the metric on $\mathbb{R}^{\omega}$ by $$D(x,y):=\sup \limits_{n\in \mathbb{N}} \bar{d}(x_n,y_n),$$ where $\bar{d}(x,y)=\min \{1,d(x,y)\}$ and $d(x,y)=|x-y|$. Then it is not so difficult to check that $D$ is the metric on $\mathbb{R}^{\omega}$.
Let's take the basis element of the uniform topology, namely $B_D(0,1)$ open ball centered at zero with radius 1. Consider $f^{-1}(B_D(0,1))$. Let's show that $f^{-1}(B_D(0,1))=\{0\}$.
Firstly, it is obvious that $0\in f^{-1}(B_D(0,1))$.
Secondly, if $t\in f^{-1}(B_D(0,1))$ $\Rightarrow$ $f(t)\in B_D(0,1)$ so $D(f(t),0)<1$ $\Rightarrow$ $\sup \limits_{n\in \mathbb{N}} \bar{d}(nt,0)<1$ $\Rightarrow$ $\bar{d}(nt,0)<1$ for all $n\in \mathbb{N}$ $\Rightarrow$ $d(nt,0)<1$ for all $n$ $\Rightarrow$ $n|t|<1$ for all $n$ $\Rightarrow$ $|t|=0$ $\Rightarrow$ $t=0$.
Hence $f^{-1}(B_D(0,1))=\{0\}$ but the last is not open set in $\mathbb{R}$ with standard topology!
hence $f$ is NOT continuous.
Remark: I guess that there many questions of such type in MSE but I have spent a whole day in order to understand how to solve it. Would be very grateful if you anybody will take a look at my solution. So please do not duplicate my question.
Your solution looks fine.
The fundamental issue to understand with this question is that as $t \to 0$, the sequences $(t, 2t, 3t, \dots)$ converge to zero pointwise, but not uniformly.