Terminal objects are unique up to unique isomorphism

Since $T_2$ is terminal, every object — in particular $T_1$ — has exactly one morphism into it; call it $u : T_1 \to T_2$. Since $T_1$ is terminal, symmetrically there is exactly one $v : T_2 \to T_1$.

Consider the composite $v \circ u : T_1 \to T_1$. Because $T_1$ is terminal, there is exactly one morphism $T_1 \to T_1$, and $1_{T_1}$ is one such morphism; since $v \circ u$ is also a morphism $T_1 \to T_1$, uniqueness forces $v \circ u = 1_{T_1}$.

By the symmetric argument using the terminality of $T_2$, the composite $u \circ v : T_2 \to T_2$ must equal $1_{T_2}$: $u \circ v = 1_{T_2}$.

A morphism with a two-sided inverse is by definition an isomorphism, so $u$ is an isomorphism $T_1 \cong T_2$ with inverse $v$. It is the unique morphism $T_1 \to T_2$ because terminality of $T_2$ already says there is only one such morphism at all — $u$ was forced from the start, and it simply turned out to be invertible.