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I have found this theorem in a calculus book

We say a function $f: I to Bbb{R}$, $I$ interval of $Bbb{R}$ is connected if $forall J subseteq I$, with $J$ interval, $f(J)$ is an interval. Prove that if $f$ is connected and if $forall y in Bbb{R} , f^{-1}({y})$ is a closed set in the relative topology of $I$, then $f$ is continuous.

I have no idea where I can start, can you give me some hints?

For $x in I$ let $I_n(x) = I cap (x - frac{1}{n},x + frac{1}{n})$. This is an open neighborhood of $x$ in $I$. We have
$$(*) phantom{xx} bigcap_{n=1}^infty f(I_n(x)) = { f(x) } .$$
"$supset$" is trivial. To verify "$subset$", let $y in bigcap_{n=1}^infty f(I_n(x))$. Then there exist $x_n in I_n(x)$ such that $f(x_n) = y$. Obviously $x_n to x$. Since $f^{-1}(y)$ is closed in $I$ , we see that $x in f^{-1}(y)$, i.e. $f(x) = y$.

Let $varepsilon > 0$. Since $I_n(x)$ is an interval, also $J_n = f(I_n(x))$ is an interval. We have $J_n = langle a_n,b_n rangle$, where $langle a_n,b_n rangle$ stands for an open, half-open or closed interval such that $a_n le f(x) le b_n$. $a_n = -infty$, $b_n = infty$ is allowed. The sequence $(a_n)$ is increasing, the sequence $(b_n)$ decreasing. But $(*)$ shows that $a_n, b_n to f(x)$, hence $f(I_n(x)) = J_n subset (f(x)- varepsilon, f(x)+ varepsilon)$ for sufficiently large $n$. This means that $f$ is continuous.