Problem Any interval in $\Bbb{R}_\mathcal{U}$ is connected in the subspace topology.
Attempt With the help of comments..
First, Note that open any interval $(a,b)$ is homeomorphic with $(0,1)$. Define a map $f:(0,1)\to (a,b)$ by $f(x)=(b-a)x+a$. It is easy to see that $f$ is homeomorphism. So $(0,1)$ is open in the subspace topology.
One can also go by the way Define a map $f\colon \mathbb{R}\to (-1,1)$ given by $f(t)=t/(1+|t|)$ is a homeomorphism, so, every open interval is connected.
Now, we'll show that $[a,b]$ is connected as a subspace topology of $\Bbb{R}_\mathcal{U}$. Construct a map $g:\Bbb{R}\to [a,b]$ define by $g(x)=x$ if $x\in [a,b]$, $g(x)=a$ if $x<a$ and $g(x)=b$ if $x>b$. we can see that $g$ is continuous. So the image of connected space is connected. and we're done.
One can also think that connectedness of open interval implies connected of closed interval since $\operatorname{Cl}((a,b))=[a,b]$.
This completes the proof.
Thanks!