If $X$ is connected and if $f:X\rightarrow R$ is non-constant and continuous, then $X$ is uncountable.
Proof.
Since $f$ is non-constant there are $a,b\in X $ such that without loss of generality $f(a)
If $X$ is connected and if $f:X\rightarrow R$ is non-constant and continuous, then $X$ is uncountable.
Proof.
Since $f$ is non-constant there are $a,b\in X $ such that without loss of generality $f(a)
To see that a non-degenerate interval is uncountable, note that any open interval $(x,y)$ is in bijection with $(0,1)$ via $f:(x,y)\to (0,1)$ given by $f(t)=\frac{t-x}{y-x}$, and that $(0,1)$ is uncountable by Cantor Diagonalization hence $(x,y)$ is. Since any non-degenerate interval contains an open interval, we are done.