Suppose $f:D \to R$ is continuous with D compact. Prove that {$x:0 \le f(x) \le 1$} is compact
Since D is compact and continuous, we know it is uniformly continuous.
Let $E=$ {$x:0 \le f(x) \le 1$}, therefore E is a bounded set.
Can I say that because $D$ is compact, we know that $f(D)$ is compact. There exists $x_1,x_2 \in D$ such that $f(x_1)=infE=0$ and $f(x_2)=supE=1$. Since E is bounded and closed, E is a compact set.
Your set is $f^{-1}[0,1]$. As $f$ is continuous, inverse image of closed set is closed. So $f^{-1}[0,1]$ is a closed subset of $D$ and as closed subset of compact Hausdorff space is also compact, you will have $f^{-1}[0,1]$ is compact.
alternative: Suffices to show every infinite set in $f^{-1}[0,1]$ has a limit point in $f^{-1}[0,1]$.
Let $x_n$ be a sequence in $f^{-1}[0,1]$, as $D$ is compact $x_n$ has a sub sequence, say $x_{n_{k}}$ which converge to some point $d\in D$. suffices to show $d\in f^{-1}[0,1]$.
Now as $f$ is continuous, $f(x_{n_{k}})$ converge to $f(d)$. Also we have $x_{n_{k}} \in f^{-1}[0,1]$ which gives us $f(x_{n_{k}}) \in [0,1]$. So $f(d)$ must belong to $[0,1]$ which tells us $d\in f^{-1}[0,1]$.