Is $S^0 = \{x \in \mathbb{R} \ |\ \|x\| = 1 \}$ path connected in the trivial topology? 
Is $S^0 = \{x \in \mathbb{R} \   |\  \|x\| = 1 \}$ path connected in the trivial topology?

I'll add some definitions here for brevity

Definition (Path): Given points $x, y$ of a space $X$, a path in $X$ from $x$ to $y$ is a continuous map $f: [a, b] \to X$ of some closed interval in the real line into $X$ such that $f(a) = x$ and $f(b) = y$.
Definition (Path Connectedness): A space $X$ is said to be path connected if every pair of points can be joined by a path in $X$.


My Attempted Proof
We have clearly that $S^0 = \{-1, 1\}$. I assert that $S^0$ is path connected in the trivial topology $\mathcal{T} = \{ \emptyset, \{-1, 1\}\}$. Pick $[a, b] \subset \mathbb{R}$ such that $a \in \mathbb{Q}$ and $b \not\in \mathbb{Q}$.
If we define $f : [a, b] \subset \mathbb{R} \to \{-1, 1\}$ such that $$f(x) = \begin{cases}-1 \ \ \ \text{if} \ \ \ x \in \mathbb{Q}\\
\ \ \ 1 \ \ \ \text{if} \ \ \ x \not\in \mathbb{Q} \end{cases}$$
for $x \in [a, b]$, then $f$ is clearly continuous as $f^{-1}\left[\{-1, 1\}\right] = [a, b]$ which is open in the subsace topology. Furthermore $f$ is a path since $f(a) = -1$ and $f(b) = 1$, hence $(S^0, \mathcal{T})$ is path connected $\ \square$

Is my proof correct? This seemed like a slightly counter-intuitive example that I constructed, or perhaps it is trivial, I'm not so sure.
 A: Your proof is correct, but your example of $f$ is more complicated than is necessary.  Suppose you have any function $f:[a,b]\to S^0$.  To check that $f$ is continuous, you just need to know that $f^{-1}(\emptyset)$ and $f^{-1}(\{-1,1\})$ are both open.  But $f^{-1}(\emptyset)=\emptyset$ and $f^{-1}(\{-1,1\})=[a,b]$, no matter what $f$ is (why?).  Since $\emptyset$ and $[a,b]$ are both open in $[a,b]$, this means $f$ is automatically continuous.
So actually every map $f:[a,b]\to S^0$ is continuous (using the trivial topology on $S^0$).  Can you find a simpler example of such an $f$ that will give a path from $-1$ to $1$?
(Also, strictly speaking, to prove $S^0$ is path-connected, you need to prove not just that there is a path from $-1$ to $1$, but also that there is a path from $1$ to $-1$, from $1$ to $1$, and from $-1$ to $-1$.  Can you explain how to get a path from $1$ to $-1$ given a path from $-1$ to $1$?  And can you explain why it is trivial that there is a path from any point to itself?)
A: Yes, you are correct. Any function $\mathbb R \to S^0$ with the trivial topology on $S^0$ is automatically continuous.
Of course, the standard topology on $S^0$ is given by $\{\emptyset, \{-1\}, \{1\}, \{-1, 1\}\}$, which is not path connected.
A: Consider the two point space $X:=\{0,1\}$ with the indiscrete topology having only two open sets $ \emptyset, X$. Any map into the indiscrete topology is continuous, and so $f(0)=0$ and $f(t)=1$ for all $t \in (0,1]$ will suffice.
This is actually the so-called "universal property" of the trivial topology (every continuous map from $X \to Y$ is continuous iff $Y$ is under the trivial topology.)
A: Hi I made an edit but please ignore it now :). I think that the topology you have defined is not a topology because it does not contain the whole interval $[-1,1]$. However if you consider it to be a topology on $\{-1,1\}$ then your proof is correct.
