Image of Interval by Continuous Function is Interval/Proof 2

Proof
Let $J$ be the image of $f$.

By Subset of Real Numbers is Interval iff Connected we need to show that $J$ is connected (and hence an interval).

not.

Then there exists a separation $S \mid T$ of $J$.

Define $A = f^{-1} \sqbrk S$ and $B = f^{-1} \sqbrk T$. $A$ and $B$ are both non-empty.

Because $f$ is continuous, by Continuous iff inverse image of any open set is open we must have $A$ and $B$ open.

Because $S \mid T$ is a separation:
 * $A \cap B = f^{-1} \sqbrk S \cap f^{-1} \sqbrk T = f^{-1} \sqbrk {S \cap T} = \O$, because $S \mid T$ is a separation.

Also, $A \cup B = f^{-1} \sqbrk S \cup f^{-1} \sqbrk T = f^{-1} \sqbrk {S \cup T} = f^{-1}(J) = I$ ($S \mid T$ is a separation of $J$).

Hence $A \mid B$ is a separation of $I$.

$I$ can certainly not be an interval (because it is not connected).

This is a contradiction.

Thus $J$ must be an interval.