What we need to show, $ab=1\bmod k$ or $(ab)\bmod n=1\bmod k$? For proving $U_k(n)≤U(n)$ I need to show that,

For each divisor $k$ of $n$, $U_k(n)$ is subgroup of $U(n)$ where, $U_k(n)=\{x\in U(n) : x=1\bmod k\}$

My attempt: as $U(n)$ is finite group for each $n\in\mathbb{Z}^+$.
Hence we may use, "finite subgroup test" to show that, $U_k(n)$  is subgroup of $U(n)$.
Clearly, $1\in U_k(n)$ since $1\equiv 1\bmod k$.
Let $a,b\in U_k(n)$ then $a=1\bmod k$ and $b=1\bmod k$, then clearly, $ab=1\bmod k$ so that $ab\in U_k(n)$ and so by finite subgroup test, $U_k(n)$ is subgroup of $U(n)$.
But, where we used $k$ divides $n$? Further, is we computed $ab$ under the operation multiplication modulo $n$? since by finite subgroup test, we need to show $U_k(n)$ is **closed under the operation of $U(n)$. Hence,  I think we need to show $(ab)\bmod n=1\bmod k$ Am I correct?
(If i am correct, then how to show $(ab)\bmod n=1\bmod k$)
Please help me......

I assume by $U(n)$ you mean $\mathbb{Z}/n\mathbb{Z}$?
We consider the example n=16 and k=3, than for $a \in \mathbb{Z}/n\mathbb{Z}$ we can not say what $a \mod k$ is, since we can choose another representant. In our example let $a=[4]$ and clearly $4 \mod 3 = 1 \mod 3$, but if we use another representant, say $[4]=[20]$, than $20 \mod 3 \neq 1 \mod 3$. This is where you use that $k$ divides $n$.
