Consider a fixed set of finite discrete symbols $\mathcal{A}$. Equip $\mathcal{A}$ wit the discrete topology which we denote by $\theta$, and $\mathcal{A}^{\mathbb{Z}^d}$ with the product topology, denoted by $\tau$.
Is then $(\mathcal{A}^{\mathbb{Z}^d},\tau)$ a compact metric space?
I think this has something to do with Tychonoff.
It's not a metric space, because you haven't defined a metric. It is, however, a metrizable space: there are metrics on it that generate the product topology. It is homeomorphic to the middle-thirds Cantor set.
One compatible metric is obtained as follows. Let $\varphi:\Bbb N\to\Bbb Z^d$ be any bijection. Let $X=\mathcal{A}^{\Bbb Z^d}$. If $x$ and $y$ are distinct points of $X$, let
$$\delta(x,y)=\min\left\{n\in\Bbb N:x\big(\varphi(n)\big)\ne y\big(\varphi(n)\big)\right\}\;.$$
Then for $x,y\in X$ set
$$d(x,y)=\begin{cases} 0,&\text{if }x=y\\ 2^{-\delta(x,y)},&\text{if }x\ne y\;; \end{cases}$$
$d$ is a metric on $X$ that generates the topology $\tau$.