Why is a finite integral domain always field?

Remember that cancellation holds in domains. That is, if $c \neq 0$, then $ac = bc$ implies $a=b$. So, given $x$, consider $x, x^2, x^3,......$. Out of finiteness there would be a repetition sometime: $x^n = x^m$ for some $n >m$. Then, by cancellation, $x^{n-m} =1$, and $x$ has an inverse.

Your proof is completable. Put $\rm\:u = x_j\ne 0.\:$ Either $\rm\:u^2 = u\:\ (so\:\ u = 1)\: $ or $\rm\: u^2 = x_{\:k}\mid 1\:$ so $\rm\:u\mid 1.\:$ Therefore all nonzero elements of $\rm\:R\:$ are units. $\:$ (note $\rm\ u^2 \ne 0\:$ by $\rm\:u\ne 0\:$). $\ $ QED

In fact one can generalize such pigeonhole-based ideas. The Theorem below is one simple way. Note that the above proof is just the special case when $\rm\:R\:$ is a domain and $\rm\:|\cal N|$ $ = 1\:.$

Theorem $\ $ If all but finitely many elements of a ring $\rm\:R\:$ are units or zero-divisors (incuding $0$), then all elements of $\rm\:R\:$ are units or zero-divisors.

Proof $\ $ Suppose the finite set $\rm\:\cal N\:$ of nonunit non-zero-divisors is nonempty. Let $\rm\: r\in \cal N.\,$ Then all positive powers $\rm\:r^n\:$ are also in $\rm\:\cal N\:$ since powering preserves the property of being a nonunit and non-zero-divisor (if $\rm\ a\,r^n = 0\:$ then, since non-zerodivisors are cancellable, we deduce $\rm\:a = 0\:$ by cancelling the $\rm\:n\:$ factors of $\rm\:r).\,$ So pigeonholing the powers $\rm\:r^n\:$ into the finite set $\rm\,\cal N$ yields $\rm\:m>n\:$ such that $\rm\:r^m = r^n,\ $ so $\,\rm\:r^n(r^{m-n} - 1) = 0\:.\:$ As $\rm r^n\in\cal N$ it is not a zero-divisor so we can cancel it, which, finally, yields that $\rm\:r^{m-n}=1,\:$ so $\rm\:r\:$ is a unit, contradiction. $\ $ QED

Corollary $\ $ Every element of a finite ring is either a unit or a zero-divisor (including $0$).
Therefore a finite integral domain is a field.

For a less trivial example see my proof here that generalizes (to "fewunit" rings) Euclid's classic constructive proof that there are infinitely many primes. Such ideas generalize to monoids and will come to the fore when one learns algebraic local-global methods, esp. localization of rings.

It sufficies to prove that there exists $1\in R$ such that $a1=1a=a$ for any $a\in R$, and that every $a\neq 0$ is invertible in $R$. So let $R=\{a_1,\dots,a_n\}$ with the $a_i$'s pairwise distinct. Let $a=a_k\neq 0$. Then the elements $$aa_1,aa_2,\dots, aa_n$$ are also pairwise distinct (if $aa_i=aa_j$ with $i\neq j$ then $a(a_i-a_j)=0$ wich forces $a_i=a_j$ since we are in an integral domain and $a\neq 0$). But then the map $\Psi:R\to R$ defined by $$\Psi(a_i)=aa_i$$ is injective by what we have proved before. Since $R$ is finite it is also surjective, then it is a bijection. This means that every element of $R$ can be written as $aa_i$ for some element $a_i\in R$. In particular $a$ itself can be written in this way: there exixsts $a_{i_0}\in R$ such that $a=aa_{i_0}=a_{i_0}a$.

Now we claim that $a_{i_0}$ is the unit element of $R$: indeed let $x=aa_i$ any element in $R$. Then $$x=aa_i=(aa_{i_0})a_i=(a_{i_0}a)a_i=a_{i_0}(aa_i)=a_{i_0}x$$ and also $$x=a_ia=a_i(aa_{i_0})=(a_ia)a_{i_0}=xa_{i_0}.$$ We shall denote this element $a_{i_0}$ with $1$. Now, from the fact that $1$ is in $R$, $1$ can be written as $1=aa_j$ for some $a_j\in R$. But then $a$ is invertible in $R$.