Show that the Stone–Čech compactification $\beta \mathbb{Z}$ is not metrizable.

In order to show that $\beta(\mathbb Z)$ is not first countable, it will suffice to show that $|\beta(\mathbb Z)|\gt2^{\aleph_0},$ since a Hausdorff space which is separable and first countable has cardinality at most $2^{\aleph_0}$ (each point is the limit of a convergent sequence of points in a countable dense set).

The space $C=\{0,1\}^\mathbb R$ is a separable compact Hausdorff space. Define a countable dense subset $S\subseteq C$ and a surjection $f:\mathbb Z\to S.$ Since $\mathbb Z$ is discrete, $f$ is continuous, and therefore extends to a continuous surjection $g:\beta(\mathbb Z)\to C,$ showing that $|\beta(\mathbb Z)|\ge|C|=2^{2^{\aleph_0}}\gt2^{\aleph_0}.$

Suppose $\beta \mathbb Z$ is metrizable. Then, since $\mathbb Z$ is not compact, there is an $b\in \beta \mathbb Z\setminus \mathbb Z$ and since $\mathbb Z$ is dense in $\beta \mathbb Z$, there is a sequence of integers $X=\left \{ x_i \right \}_{i\in \mathbb N}$ that converges to $b$.

Now take any two disjoint subsequences $X_1$ and $X_2$ of $X$ and note that they are closed in $\mathbb Z$. As $\mathbb Z$ inherits the metric on $\beta \mathbb Z$, it is a normal space. Therefore, there is a continuous $f:\mathbb Z\to [0,1]$ s.t.$f(X_1)=0$ and $f(X_2)=1$, which extends continuously to a $g:\beta \mathbb Z\to [0,1]$.

But $X_1$ and $X_2$ both converge to $b$ and as $g$ is continuous we must have $g(b)=g(\lim X_1)=\lim g(X_1)=\lim f(X_1)=0$ and also $g(b)=g(\lim X_2)=\lim f(X_2)=1$, which is a contradiction.

A compact metric space is separable.

Take an uncountable family of subsets of $\mathbb Z$, any two of which have finite intersection. Then their closures intersected with the "remainder" $\beta \mathbb Z \setminus \mathbb Z$ give you an uncountable family of pairwise disjoint open sets in the compact space $\beta \mathbb Z \setminus \mathbb Z$. So $\beta \mathbb Z \setminus \mathbb Z$ is not metrizable, and therefore of course neither is $\beta \mathbb Z$.