This is Fact 1.4 in Tom Leinster's informal introduction to topos theory. It states the following: if there exists a mono $t:T \hookrightarrow \Omega$ that classifies the monos in our category, in the sense that for any mono $A \hookrightarrow X$ there exists a unique pullback square $(A \to T \hookrightarrow \Omega) = (A \hookrightarrow X \to \Omega)$, then the object $T$ is necessarily terminal.
I see how to construct a morphism from any object $Y$ to $T;$ namely, the identity $Y \to Y$ is a mono and so induces a morphism $Y \to T$ sitting inside a pullback square. However, I don't see how to use the extra information to conclude that this is the only morphism from $Y$ to $T.$
As you say, there is a morphism $f : Y \to T$ making the following diagram a pullback square, $$\require{AMScd} \begin{CD} Y @>{f}>> T \\ @| @VV{t}V \\ Y @>>{t \circ f}> \Omega \end{CD}$$ and $f : Y \to T$ is the unique morphism with this property (because $t : T \to \Omega$ is a monomorphism). So we need to show that, for any morphism $g : Y \to T$, we have $f = g$. But it is straightforward to check that $$\begin{CD} Y @>{g}>> T \\ @| @VV{t}V \\ Y @>>{t \circ g}> \Omega \end{CD}$$ is always a pullback square, so indeed $f = g$.