First recall that a (potentially unbounded) operator $T:D(T)\subseteq X\to Y$ is closed whenever $(x_n)_{n=1}^\infty\subset D(T)$ convergent to $x\in X_0$ and $(Tx_n)_{n=1}^\infty$ convergent in $X_1$ together imply that $x\in D(T)$ and $Tx=\lim_{n\to\infty}Tx_n$.
On the Wikipedia page for unbounded operators, differentiation $\frac{d}{dx}:C^1([0,1])\to C([0,1])$ is given as an example of a closed, unbounded operator. In this example, $X,Y$ are both taken to be $C([a,b])$ and $D(T)=C^1([a,b])$. As our unbounded operator we have $T=\frac{d}{dx}$.
Whilst I have no trouble seeing that in this example $T$ is unbounded, I am not so convinced that $T$ is closed. In particular, I know that $C^1([0,1])$ is dense in $C([0,1])$, but not closed; so it isn't so obvious that a convergent sequence $f_n\in C^1([0,1])$ should have limit $f\in C^1([0,1])$. How, exactly, should one proceed in trying to show this?
To check that $\frac{d}{dx}$ is closed, you should check that if $f_n \in C^1([0,1])$ and $f_n \to f$, $f_n' \to g$ in $C([0,1])$, then $f \in C^1([0,1])$ and $f' = g$.
By the fundamental theorem of calculus, we can write $f_n(x) = f_n(0) + \int_0^x f_n'(s) ds$. In particular, by sending $n \to \infty$, $$f(x) = f(0) + \int_0^x g(s) ds$$ so $f \in C^1([0,1])$ and $f' = g$.