This is really a comment on t3suji's answer, but it's too long to be a comment as such.

t3suji's answer is the canonical one in the following precise sense. Let $e: X \to Y$ be a morphism in any category. It's an elementary exercise to show that the following conditions on $e$ are equivalent:

$e$ is an epimorphism

the square
$$
\begin{array}{ccc}
X &\stackrel{e}{\to} &Y \\
e\downarrow & &\downarrow 1_Y \\
Y &\stackrel{1_Y}{\to} &Y
\end{array}
$$
is a pushout

for some morphism $f: Y \to Z$, the square
$$
\begin{array}{ccc}
X &\stackrel{e}{\to} &Y \\
e\downarrow & &\downarrow f \\
Y &\stackrel{f}{\to} &Z
\end{array}
$$
is a pushout.

I'll only use the equivalence 1 $\iff$ 3 here. The other implications are just scene-setting.

Suppose we want to show that a particular morphism $e$ is *not* epi. Assuming that there are enough pushouts around, we can argue as follows. Form the pushout square
$$
\begin{array}{ccc}
X &\stackrel{e}{\to} &Y \\
e\downarrow & &\downarrow f \\
Y &\stackrel{g}{\to} &Z.
\end{array}
$$
If $f \neq g$ then the implication 1 $\Rightarrow$ 3 tells us that $e$ is not epi. Moreover, this strategy is bound to work, in the sense that if $f = g$ then the implication 3 $\Rightarrow$ 1 tells us that $e$ is epi after all.

It only remains to see that this is indeed what t3suji did. In his/her situation, $e$ was the inclusion $X \to Y$. He/she then took the coequalizer of the two obvious maps $X \to Y + Y$ (where $+$ means coproduct, i.e. disjoint union). For elementary and totally general reasons, this is the same thing as taking the pushout just mentioned. The morphisms that t3suji called $\iota_1$ and $\iota_2$, I called $f$ and $g$. Finally, although t3suji's pushout is in the category of *all* topological spaces, he/she then verified that the space $Z$ is indeed Hausdorff, from which it follows that it's also a pushout in Hausdorff spaces.

So now you know, in principle, how to answer any question of the form "prove that such-and-such a morphism isn't epi".