Vtyjkyuk

Proving, using the definition of a limit that the function is continuous everywhere
where $z,a in C$

$$f(z) = operatornameRe(z) $$

$f:S$ $subset C to C$

So I got to $|operatornameRe(z) - operatornameRe(a)| = 0.5|z - a + bar z - bar a|$
but I don't know how I can use this expression to show that if $|z - a| < delta$ then $|operatornameRe(z) - operatornameRe(a)| < varepsilon$

for any $varepsilon > 0$

Let $z=(x+iy)$ and $w=(a+ib)$

Note that your function is $$ f(x+iy)=x$$and $$ f(a+ib)=a$$

$$ |f(z)-f(w)|=|f(x+iy)-f(a+ib)|=|x-a|le sqrt (x-a)^2+(y-b)^2=|z-w|$$

For a given $epsilon >0$ let $delta = epsilon $

If $$|z-w|<delta$$ then $$ |f(z)-f(w)|le |z-w|<delta =epsilon$$

Thus $ f $ is continuous.

It is easy to show that $z_n to z$ implies $overlinez_n to overlinez$. Hence, $z_n to z$ implies $$operatornameRe z_n = z_n + overlinez_n to z + overlinez = operatornameRe z.$$
This shows that $operatornameRe(cdot)$ is continuous.

Forgot about the Triangle Inequality
$|operatornameRe(z) - operatornameRe(a)| = 0.5|z - a + bar z - bar a| leq 0.5(|z - a| + |bar z - bar a|) = 0.5(2|z - a|) = |z-a|$ using triangle inequality

then pick $delta = varepsilon$

Therefore if $|z - a| < delta$

$|operatornameRe(z) - operatornameRe(a)| < delta = varepsilon$

Thanks the guy in the comments