Show the quotient by equivalence $A \sim B \iff A^2+23A=B^2+23B$ is Hausdorff So, in the set $T$ of 2x2 uppertriangular matrices one has defined
$$
A \sim B \iff A^2+23A=B^2+23B
$$
Is the quotient $\quad T/\sim \quad$ Hausdorff?
My attempt

*

*I've shown that every small enough neighbourhood of the matrix $A^2+23A$ is non-empty.

*Then I said that small enough neighbourhoods of two distinct matrices $A^2+23A$ and $B^2+23B$ does not intersect(I used that as an obvious fact)

So, from the last step I concluded the quotient space is Hausdorff.
Am I right? Other ideas?
 A: 
I've shown that every small enough neighbourhood of the matrix $A^2+23A$ is non-empty.

This is true, but something of a tautology. A neighborhood of a point (it doesn't need to be "small enough") is always non-empty because by definition it contains the point itself!

Then I said that small enough neighbourhoods of two distinct matrices $A^2+23A$ and $B^2+23B$ does not intersect(I used that as an obvious fact)

What you need to make this statement less handwavy is the knowledge that $T$ itself (in its role as the space where $A^2+23A$ and $B^2+23B$ live) is Hausdorff.
The general principle at work is: If you have topological spaces $X$ and $Y$ and a continuous function $f: X\to Y$ you can form the quotient $X/{\sim_f}$ where $x_1\sim_f x_2$ iff $f(x_1)=f(x_2)$. This quotient is in bijective correspondence with $f(X)\subseteq Y$.
Furthermore the topology of $X/{\sim_f}$ is at least as fine as the subspace topology on $f(X)$ -- since preimages of open sets in $Y$ under the continuous map $f$ are open in $X$.
Now if $Y$ is Hausdorff, then every subspace of it is Hausdorff too, and therefore $X/{\sim}_f$ is Hausdorff because making a topology finer will not harm the Hausdorff property.
