Vtyjkyuk

In the following (Terence 2006, p.38) I have to prove as exercice:

Let $f : X to Y$ be a function from a connected metric space $(X, d)$ to a metric space $(Y, d_disc )$ with the discrete metric. Show that $f$ is continuous if and only if it is constant

I have proven this proposition without using the fact that $X$ is connected, so obviously I did something wrong. I would be very gratefull if you could point out my mistake.

Proof: $impliedby$ Suppose $f$ is constant, i.e.,
$$forall x,x'in X: f(x) = f(x')$$
Choose an arbitrary $varepsilon>0$. Then, no matter what our choice of $delta_varepsilon$ is, due to the definition of the discrete metric we always have
$$ d(x,x_0)<delta_varepsilon implies d_disc(f(x),f(x_0))=0<varepsilon$$

$implies$ Now suppose that $f$ is continuous. Then, for any $x_0 in x$ we have
$$forall varepsilon>0 quad exists delta_varepsilon>0 quad forall x in X: quad d(x,x_0)<delta_varepsilon implies d_disc(f(x),f(x_0))<varepsilon$$
Now suppose, for the sake of contradiction, that $f$ is not constant, i.e.,
$$exists x_0 in X quad forall x in X/x_0: f(x) neq f(x_0)$$
But then, for such $x_0$ we cannot find a $delta_0.5$ so that $d(x,x_0)<delta_varepsilon implies d_disc(f(x),f(x_0))<0.5$ - a contradiction.

Let $y_1,y_2 in f(X)$. Write $Y=Y_1 cup Y_2$ where $Y_1 cap Y_2 = emptyset$ and $y_1 in Y_1,y_2 in Y_2$. Since $Y_1$ and $Y_2$ are open sets as its a discrete metric space, $f^-1(Y_i)$ must be open and $X=f^-1(Y_1) cup f^-1(Y_2)$ contradicting the connectedness of the metric space $X$.