Is my proof that $U_{pq}$ is not cyclic if $p$ and $q$ are distinct odd primes correct?

Prove that $U_{pq}$ is not cyclic if $p$ and $q$ are distinct odd primes.

I am a self taught person. I just learned this and tried this on my own and came up with this.

$x \equiv 1 \pmod{p}$ and $x \equiv -1 \pmod{q}$ has a solution $[a]_{pq}$, since $p$ and $q$ are relatively prime. Because $q$ is an odd prime, $[-1]_{pq}$ is not a solution, so $[a]_{pq}\neq [-1]_{pq}$. But $a^2 \equiv 1 \pmod{p}$ and $a^2 \equiv 1 \pmod{q}$, so $a^2 \equiv 1\pmod{pq}$ since $p$ and $q$ are relatively prime, and thus $[a]_{pq}$ has order 2.

Can someone please tell me if this proof this correct? Please help with proof as I learned it just now as a self taught person.

This proof is indeed correct. You have produced an element of order $2$ which is not $-1$--this proves the result as Andre Nicolas said.

Phrasing what you said differently, note that the Chinese Remainder Theorem gives you a ring isomorphism $\def\Z{\mathbb{Z}}$

$$\mathbb{Z}/pq\Z\cong (\Z/p\Z)\times(\Z/q\Z)$$

This gives rise to an isomorphism of unit groups:

$$(\Z/pq\Z)^\times\cong (\Z/p\Z)^\times\times(\Z/q\Z)^\times$$

Now, $(\Z/p\Z)^\times$ and $(\Z/q\Z)^\times$ are both abelian groups of order $p-1$ and $q-1$. From here you can observe (just as you did!) that we have produced two distinct elements of order $2$, or you can note that the product of two finite abelian groups of non-coprime orders is never cyclic by FTFGAG.

We prove the lemma needed to finish your argument.

Suppose that $A$ is a cyclic group of order $n$, where $n$ is even. Let $g$ be a generator of $A$. Then $g^{n/2}$ is the only element of order $2$.

For suppose that $g^k$ has order $2$, where $1\le k\le n-1$. Then $g^{2k}$ is the identity. It follows that $n$ divides $2k$, which forces $k=n/2$.

Remark: In a similar way, one can prove the following useful result. Let $A$ be an abelian group, and suppose that $a$ has finite order $d$. Then $a^k$ has order equal to $\frac{d}{\gcd(k,d)}$.