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I have to use the following definition to prove that the function Re $z$ is nowhere differentiable:
Let $f$ be a complex-valued function defined in a neighborhood of $z_0$. Then the derivative of $f$ at $z_0$ is given by
$$\frac{df}{dz}(z_0)=f'(z_0)=\lim_{\Delta z \to 0} \frac{f(z_0+ \Delta z)-f(z_0)}{\Delta z},$$ provided this limit exists. (Such an $f$ is said to be differentiable at $z_0$)

First I need to understand what Re $z$ means? This is just the real part of $z$, where $z$ is $x+yi$, so the real part is just $x$, but I don't understand how the the real part alone is considered a function?

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    As an example: if $f(z)=\operatorname{Re} z$, then $f(3+5i)=3$ and $f(2-6i)=2$. – 2017-02-20
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    For each input $z=x+yi$, there exists precisely one output $\operatorname{Re}(z)=x$. That's the very definition of a function: for each input there's one output. – 2017-02-20
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    For each $z$ the real part is $x$ so each input $z$ produces a unique output, so it is a function. Try using the Cauchy-Riemann equations. – 2017-02-20
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    It's the function $x + yi \mapsto x + 0i$. – 2017-02-20

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