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Let $f:X\to Y$ be a homeomorphism with continous inverse $g:Y\to X$. I want to show that $U$ is a neighbourhood of $x\in X$ iff $f(U)$ is a neighbourhood of $f(x)$.

One direction is obvious, just apply $g$ to $f(U)$ and $f(x)$ and by continuity this yields the result. However, I fail to show the other direction because I don't see how I can use the continuity of $f$ on $x$. Thanks for help with this step!

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    I feel like your "one direction" is not as obvious as you think. If $A$ is a neighbourhood of $a$, and $h$ is a continuous function, then $h(A)$ need not be a neighbourhood of $h(a)$. – 2017-01-17
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    That's exactly the direction I find not obvious. What is obvious is that if $f(U)$ is a neighbourhood of $f(x)$ then $gf(U)=f^{-1}f(U)=U$ is of $x$. This is the definition of $f$ being continous and uses that it is bijective. The other direction, the one you gave, is not clear to me as well. – 2017-01-17
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    Hint: $U = g(f(U))$, $x = g(f(x))$. – 2017-01-17
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    Ok, now I see how I can apply $f$ to $x$ and use continuity. This is a nice solution, thank you! – 2017-01-17

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An homeomorphism is a continuous AND an open map, it means for all open $U$ in $X$ we must have $f(U)$ open in $Y$, and that for all $V$ open in $Y$ we must have $f^{-1}(V)$ open in $X$.

If $U$ is a neighborhood of $x \in X$, must contain an open set $G$ (in $X$) such that $x \in G \subset U$. Since $f^{-1}$ is continuous then $f(G)$ is open (in $Y$). Since $f(x) \in f(G) \subset f(U)$, then $f(U)$ is a neighborhood of $f(x)$.

If $V$ is a neighborhood of $f(x)$, it must contain an open set $H$ (in $Y$) such that $f(x) \in H \subset V$. Since $f$ is continuous then $f^{-1}(H)$ is open (in $X$) and satisfies $x \in f^{-1}(H) \subset f^{-1}(V)$, so $f^{-1}(V)$ is a neighborhood of $x$.

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    True, but I do not just ask for open sets but for $f(U)$ being a neighbourhood of $f(x)$. The assertion you wrote would also be correct for just open maps but I don't feel that they necessarily preserve neighbourhoods. – 2017-01-17
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    Being a neighborhood means it contain an open set that contains the point. – 2017-01-17
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    @positrono802 in line 3,4 don´t You mean:Since $f$ is open $f(G)$ is open in $Y$? – 2017-01-17
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    The thing I was puzzled with is the "contains"-part. Now I your edited answer I see the inclusion of sets. Thanks a lot! – 2017-01-17
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    @PeterMelech You're right. Fixed. – 2017-01-17