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Universal property of quotient spaces

A typical textbook theorem about quotient space is as follows:
>**Theorem**: Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X / \sim ightarrow Y$ such that $f=g{\circ} \pi$.
It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces) because $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$ for any open set $V$ in $Y$ with the well-defined $g([x]):=f(x)$.
But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which is supposed to be easy too. But I got stuck.
Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/~$ such that $g(O)\subset U$.
Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
**Question**: How can I go on from here or do I need to change the argument completely?
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I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.

A typical textbook theorem about quotient space is as follows:
>**Theorem**: Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X/{\sim} ightarrow Y$ such that $f=g \circ \pi$.
It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces) because $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$ for any open set $V$ in $Y$ with the well-defined $g([x]):=f(x)$.
But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which is supposed to be easy too. But I got stuck.
Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/~$ such that $g(O)\subset U$.
Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
**Question**: How can I go on from here or do I need to change the argument completely?
---
I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.

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