To prove that $X$ is compact, it suffice to show that it is totally bounded and complete.
To show that it is totally bounded, let $\epsilon\gt 0$. Let $m\gt 0$ be such that $\frac{1}{m}\lt\epsilon$. Now, there are only $n^m$ possible finite sequences of length $m$ with values in $\{1,2,\ldots,n\}$. For each such sequence $s$, let $\mathcal{O}_s$ consist of all elements of $X$ that agree with $s$ in the first $m$ points. The set is open, since for every $(x_n)\in \mathcal{O}_s$, any sequence within $\frac{1}{m}$ of $(x_n)$ must agree with $x_n$ on the first $m$ terms, and hence have $s$ as its initial term. Moreover, the distance between any two elements of the set is less than $\epsilon$, so the radius of $\mathcal{O}_s$ is less than $\epsilon$. Since $X$ is the union of the $\mathcal{O}_s$, we see that $X$ is totally bounded.
Now suppose that $(x_n)_m$ is a Cauchy sequence. Then for every $k\in\mathbb{N}$ there exists $M(k)\gt 0$ such that if $j,\ell\geq M(k)$, then $d((x_n)_i,(x_n)_j)\lt \frac{1}{k}$. That means that $(x_n)_i$ and $(x_n)_j$ agree in the first $k$ terms. Define the sequence $(y_n)$ by letting $y_k$ be the common $k$th term to all sequences with index greater than $M(k)$. Show that $(x_n)_m\to (y_n)$.
Thus, $X$ is compact.