I am new on this topology stuff, so came to ask for help, my intuition tells me that I need to use a topological invariant to see when they are homeomorphic but I am not quite sure how to proceed.
Prove that $\Bbb R \cong \Bbb {R^n}$ if and only if $n=1$ and $\Bbb {S^1} \cong \Bbb {S^n}$ if and only if $n=1$
A more elementary way than what Chris Custer suggests is this: if we remove a point from $\mathbb{R}$, we get a disconnected set. But if $n>1$, $\mathbb{R}^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $\mathbb{S}^n$ minus one point is homeomorphic to $\mathbb{R}^n$.
Suppose there is such homeomorphism $f:\Bbb R\to\Bbb R^n$, then setting $A:=(-\infty,0)\cup(0,\infty)$ it is enough to check that $\Bbb R^n\setminus \{f(0)\}$ is connected for $n>1$ but $A$ is disconnected, hence $f^{-1}$ cannot be continuous, so it is not an homeomorphism.
For the other case you can proceed analogously noticing that $\Bbb S^n\setminus\{a\}\cong\Bbb R^n$ via stereographic projection for any chosen $a\in\Bbb S^n$.
One nice way is to use the invariant of homology. $H_n(S^n)\cong\Bbb Z$ and $H_m(S^n)\cong0\,,m\neq n,0$.
$\Bbb R^n$ can then be done using the one-point compactification.