Let $\forall n\geq 2, \;P(n)$ be the statement that $\exists \; t\in N $ such that $1+2+\ldots t < n \leq 1+2+3\ldots (1+t)$
that is, $P(n)$ is true if $\exists \; t\in N $ such that $\frac{(t)(t+1)}{2} < n \leq \frac{(t+1)(t+2)}{2}$
$P(2)$ is true take $t=1$
Assume $P(n)$ is true $\rightarrow \exists\; t_{n} $ such that
$$\frac{(t_n)(t_n+1)}{2} < n \leq \frac{(t_n+1)(t_n+2)}{2}$$
We are looking for $t_{n+1}$ that satisfies $P(n+1)$
Case $1$: $$n<\frac{(t_n+1)(t_n+2)}{2}$$
in this case $t_{k+1}=t_k$ does the job since
$$\frac{(t_n)(t_n+1)}{2} < n+1 \leq \frac{(t_n+1)(t_n+2)}{2}$$
Case $2$: $$n=\frac{(t_n+1)(t_n+2)}{2}\rightarrow \frac{(t_n+1)(t_n+2)}{2} < n+1 \leq \frac{(t_n+1)(t_n+2)}{2} + (t_n +2) $$
$$\frac{(t_n+1)(t_n+2)}{2} < n+1\leq \frac{(t_n+2)(t_n+3)}{2}$$
so in this case $t_{n+1}=t_n + 1$ does the job.
Now on the uniqueness part:
Suppose, $\exists t_1, t_2 \in N$ such that for some $n\geq 2$ we have,
$$\frac{(t_i)(t_i+1)}{2} < n \leq \frac{(t_i+1)(t_i+2)}{2} \; \; i\in [2]$$
Want to show : $t_1 = t_2$
Due to ordering in $N$, if $t_1 \neq t_2 \rightarrow t_1 < t_2 $ or $t_2 < t_1$
WLOG, assume $t_1<t_2\rightarrow t_2 = t_1 + k \; \text{for some} \; k\in N$
Need help in proceeding from here
Will be better to include uniqueness of $t$ in $P(n)$ itself?
Inducting on $n$ isn't your best option. It's easier to prove by induction on $t$ that $T_t:=\sum_{k=1}^tk$ is a strictly increasing sequence in $\Bbb N$. Since any such sequence contains arbitrarily large elements (in particular $T_t\ge t$), for any $n\in\Bbb N$ there is at least one $t\in\Bbb N$ with $n\le T_{t+1}$. Therefore, there is a least such $t$, whence $n\not\le T_{t-1+1}\implies T_t<n\le T_{t+1}$.