I've been given a theorem:
Let $X$ be a space, ${\sim}$ be an equivalence relation on $X$ and $X/{\sim}$ the quotient space.
Suppose $f:X\rightarrow Y$ is a quotient map and $x \sim y\Leftrightarrow f\left(x\right)=f\left(y\right)$. Then, $X/{\sim}$ is homeomorphic to $Y$.
Here is my proof so far:
Let $\left[x\right]\in X/{\sim}$ denote the equivalence class of $x\in X$
Let $g:X\rightarrow X/{\sim}$, $x\mapsto\left[x\right]$ be the quotient function
Define $\widetilde{f}:X/{\sim}\rightarrow Y$, $\widetilde{f}\left(\left[x\right]\right)=f\left(x\right)$ so that $f=\widetilde{f}\circ g$
Then, $\widetilde{f}$ is well-defined and a bijection*
Moreover, $U$ is open in $X/{\sim}\Leftrightarrow g^{-1}\left(U\right)$ is open in $Y$
$\Leftrightarrow f^{-1}fg^{-1}\left(U\right)$ is open in $Y$
$\Leftrightarrow fg^{-1}\left(U\right)$ is open in $Y$
$\Leftrightarrow\widetilde{f}gg^{-1}\left(U\right)$ is open in $Y$
$\Leftrightarrow\widetilde{f}\left(U\right)$ is open in $Y$
Therefore, $\widetilde{f}$ is a homeomorphism
I'm not sure how to show that $\tilde{f}$ is a well-defined bijection.
$\newcommand{\s}{{\sim}}\newcommand{\t}[1]{\widetilde{#1}\!}$I assume that your issue is in showing the "well-defined" part.
Note that you are defining $\t{f}([x]) = f(x)$. To show that this is well-defined, you need to show that $f(x)$ "only depends on $[x]$ and not $x$".
More precisely, you need to show that: $[x] = [y] \implies f(x) = f(y)$.
But this follows precisely from your definition of $\s$. Indeed, you have $$[x] = [y] \implies x \sim y \implies f(x) = f(y).$$
(In fact, note that the above implications are reversible. Thus, $[x] = [y] \iff f(x) = f(y)$. This also shows you that $\t{f}$ is an injection. That it is a surjection follows from the fact that $f$ was a surjection to begin with.)