Here's a theorem in the course notes of a course on Polish groups:
Let $X$ and $Y$ be Polish spaces and $f:X\to Y$ continuous.
$f(X)$ has the Baire property.
In the course note, it's written that this follows from the existence of $U(A)$ (which I define bellow). The proof should be easy otherwise I would have written it. Unfortunately, while reviewing the notes, I forgot the proof and I'm unable to prove the theorem above.
Here's the definition of $U(A)$:
Let $X$ be a Polish space and $A\subseteq X$.
$$U(A):=\bigcup\{O\mid A\text{ is comeager in $O$, i.e, $O\backslash A$ is meager}\}.$$
We have the following facts:
$U(A)$ is open.
$A$ is comeager in $U(A)$.
$A$ has the Baire property $\iff$ $A\backslash U(A)$ is meager.
Attempt
So I tried to show that $f(X)\backslash U(f(X))$ is meager.
If $f(X)$ is meager, we're done. Otherwise $U(f(X))\neq\emptyset$ and $U(f(X))\backslash f(X)$ is meager. I coudln't find a way to use the continuity of $f$ to conclude. I also tried to prove the theorem by contradiction using $$f(X)\backslash U(f(X))=\bigcap\{f(X)\backslash O\mid f(X)\text{ is comeager in $O$}\}$$
but I couldn't get to the result. Could you please help me?