Suppose that $f$ is not surjective, then $f(U) \subset U \setminus \{u\}$ for some $u$. It's a well known fact that $U \setminus \{u\}$ is homeomorphic to $(0,1)$. $f(U)$ is a connected subset of $(0,1)$ (because $U$ is connected), thus it's an interval $(a,b)$ with $a<b$. Let $c \in (a,b)$ and $v \in U$ such that $f(v) = c$. Then by injectivity $f(U \setminus \{v\}) = (a,c) \cup (c,b)$. But $U \setminus \{v\}$ is connected, while $(a,c) \cup (c,b)$ isn't, a contradiction.
Hint:
$$t\overset{\varphi}\mapsto{\displaystyle \left({\frac {2t}{1+t^{2}}},{\frac {t^2-1}{1+t^{2}}}\right)}$$
is the inverse of your map from $\mathbb{R} \to S^1\setminus\{(0,1)\}$. This will prove that your initial map is both injective and surjective. Note that at $x=\infty$, we get $\varphi(\infty)=(0,1)$ but $\infty\not\in\mathbb{R}$ of course.
If you want to see how these two maps are obtained, assuming you don't know it already, just fix a line at $(0,1)$ and change the slope to sweep $\mathbb{R}$. This will give a (bi-continuous) one-to-one correspondence between $S^1\setminus\{(0,1)\}$ and $\mathbb{R}$ as long as the slope of the line doesn't become parallel to the $x$-axis. And it never becomes parallel to the $x$-axis unless the line is tangent to the circle at $(0,1)$.
Meanwhile, note that just proving your function $\psi$ is continuous is not enough. You should also show that its inverse is continuous. In this case, $\varphi$ is given by two polynomials whose denominators never become zero. So, $\varphi$ is continuous on $\mathbb{R}$.
Addendum
More precisely, if you wonder how $\varphi$ is found, first write the equation of the line passing through $(0,1)$ with the varying slope $t$
$$y-1=t(x-0) \implies y=tx+1$$
You want to see how $t$ determines a point on the circle. So, you have to find its intersection with the unit circle. So, your intersection point must satisfy $x^2+y^2=1$. This gives
$$x^2+(tx+1)^2=1$$
Now you've found an equation which is quadratic in $t$. So, you'll find two solutions. One solution is $t=0$ which corresponds to the point $(0,1)$ where our line has been fixed to the circle, and the other solution gives us the point that we're looking for.
Best Answer
By contraddiction, if $\phi$ would be an homeomorphism, then $[0,1)\setminus \{\frac{1}{2}\}$ is homeomorphic to
$\mathbb{S}^1\setminus \{\phi(\frac{1}{2})\}$
but the first space is not connected while the second is connected.
Another way can be to observe that $[0,1)$ is not compact while $\mathbb{S}^1$ is a compact Space because is a closed and limited subset of $\mathbb{R}^2$