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I have written a proof of the proposition (that the continuous image of a connected set is connected) slightly different from the ones on here so far (of which I am aware there are many). Is the following correct?

I have a continuous function $f:X \rightarrow Y$, where $X$ and $Y$ are metric spaces such that $X$ is connected. I want to show that $f(X)$ is connected as a metric space, with the metric induced by $Y$.

Suppose, for a contradiction, that $f(X)$ is not connected. That is, there exists open sets $A,B$ such that $f(X)=A\cup B$ and $A \cap B = \emptyset$. Note that $A,B\subseteq Y$, and so, by the continuity of $f:X \rightarrow Y$, $f^{-1}(A)$ and $f^{-1}(B)$ are both open in $X$. Additionally, $f^{-1}(A)\cap f^{-1}(B)= \emptyset$ (since, otherwise, $\exists x^*$ such that $x^*\in f^{-1}(A) \cap f^{-1}(B)$ and so $f(x^*)\in A \cap B$, contradicting the assumed disjointedness of $A$ and $B$). Finally, we show that $X=f^{-1}(A)\cup f^{-1}(B)$.

$w\in f^{-1}(A)\cup f^{-1}(B) \iff f(w) \in A \cup B = X \iff w\in X$.

So $X=f^{-1}(A)\cup f^{-1}(B)$, and is not connected. A contradiction.

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    $\begingroup$ This is correct, and never uses the fact that either $X$ or $Y$ is a metric space, so you've actually proved the proposition for topological spaces in general. $\endgroup$
    – Robert Shore
    Oct 18, 2022 at 0:17
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    $\begingroup$ This looks like the “standard proof” to me. $\endgroup$
    – Ted Shifrin
    Oct 18, 2022 at 0:20
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    $\begingroup$ A minor point: you seem to be misusing $:=$. That's usually used to introduce a definition; in contrast, if you're merely asserting equality of two preexisting objects, which is the case everywhere here, then you should use plain $=$. $\endgroup$
    – Daniel Schepler
    Oct 18, 2022 at 0:22
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    $\begingroup$ $A$ and $B$ are open sets in $f(X),$ not $Y.$ You will need a little more care. $\endgroup$
    – Thomas Andrews
    Oct 18, 2022 at 0:52
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    $\begingroup$ @RobertShore No, it is not correct. $A\cup B=f(X)$ means we are talking about open sets in $f(X),$ not in $Y.$ $\endgroup$
    – Thomas Andrews
    Oct 18, 2022 at 0:57

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This is fine. I'll just make one small comment that it's totally unnecessary to phrase this as a proof by contradiction. This is a proof by contrapositive: you've shown that if $f(X)$ is not connected then $X$ is not connected.

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