Here, adapted from an example and a problem in Engelking and with lots of blanks filled in, is an example of a zero-dimensional Tikhonov space with a subspace $-$ in fact a closed subspace $-$ of dimension greater than $0$.
The first step is to construct a zero-dimensional Tikhonov space $X$ that is not strongly zero-dimensional; this construction is apparently due to Dowker.
Let $I=[0,1]$, and let $Q=I\cap\mathbb{Q}$. Define the relation $\sim$ on $I$ by $x\sim y$ iff $|x-y|\in Q$; clearly $\sim$ is an equivalence relation with countable equivalence classes, so it has $2^\omega\ge\omega_1$ equivalence classes, one of which is $Q$ itself. Note also that each $\sim$-class is dense in $I$.
Choose $\omega_1$ of these classes, not including $Q$, and enumerate them as $\{Q_\xi:\xi<\omega_1\}$. For $\eta<\omega_1$ let $S_\eta=I\setminus\bigcup_{\eta\le\xi<\omega_1}Q_\xi\;;$ each $S_\eta$ is zero-dimensional, since its complement is dense in $I$. We now use the $S_\eta$ to define some subspaces of $(\omega_1+1)\times I$:
$\begin{align*} X_\eta&=\bigcup_{\xi\le\eta}\Big(\{\xi\}\times S_\xi\Big), \qquad\eta<\omega_1\\ X&=\bigcup_{\xi<\omega_1}\Big(\{\xi\}\times S_\xi\Big)=\bigcup_{\eta<\omega_1}X_\eta\;. \end{align*}$
Note that $X_\eta=\big((\eta+1)\times I\big)\cap X$ is a clopen subset of $X$ for each $\eta<\omega_1$. Moreover, $X_\eta\subseteq (\eta+1)\times S_\eta$, which is clearly zero-dimensional, so each $X_\eta$ is zero-dimensional, and therefore so is $X$. This also ensures that $X$ is Tikhonov. (In fact $X$ is normal, but that’s a bit harder to prove.)
However, $X$ is not strongly zero-dimensional, because the closed sets $H_0=\omega_1\times\{0\}$ and $H_1=\omega_1\times\{1\}$ can’t be separated by clopen sets in $X$. To prove this, suppose that $U$ is a clopen subset of $X$ such that $H_0\subseteq U\subseteq X\setminus H_1$. Let $x\in I$ be arbitrary, and fix $\alpha<\omega_1$ such that $x\in S_\alpha$; clearly $x\in S_\eta$ whenever $\alpha\le\eta<\omega_1$.
Suppose that for every $\eta<\omega_1$ and $n\in\omega$ there are $\langle \xi_0,x_0\rangle\in U$ and $\langle \xi_1,x_1\rangle\in X\setminus U$ such that $\xi_0,\xi_1\ge\eta$, $|x-x_0|<2^{-n}$, and $|x-x_1|<2^{-n}$. Then we can recursively construct a sequence $\Big\langle\langle \xi_n,x_n\rangle:n\in\omega \Big\rangle$ such that $\xi_0\ge\alpha$, $\xi_{n+1}>\xi_n$ and $|x-x_n|<2^{-n}$ for each $n\in\omega$, $\langle \xi_n,x_n\rangle\in U$ when $n$ is even, and $\langle \xi_n,x_n\rangle\in X\setminus U$ when $n$ is odd. Let $\eta=\sup_n\xi_n$. Then $\Big\langle\langle \xi_n,x_n\rangle:n\in\omega \Big\rangle\to\langle \eta,x\rangle\in X_\eta\;,$ so $\langle \eta,x\rangle\in U\cap (X\setminus U)=\varnothing$, which is absurd. Thus, for each $x\in I$ there are $\eta(x)<\omega_1$ and $n(x)\in\omega$ such that $B(x)\triangleq\big\{\langle \xi,y\rangle\in X:\xi\ge\eta(x)\land|y-x|<2^{-n(x)}\big\}$ is a subset either of $U$ or of $X\setminus U$.
Let $I_0=\{x\in I:B(x)\subseteq U\}$ and $I_1=I\setminus I_0=\{x\in I:B(x)\subseteq X\setminus U\}$. Clearly $I_0\cap I_1=\varnothing$, $0\in I_0$, and $1\in I_1$, so $\{I_0,I_1\}$ is a partition of $I$. But it’s easy to see that if $x\in I_i$, then $\{y\in I:|y-x|<2^{-n(x)}\}\subseteq I_i$ as well, so $I_0$ and $I_1$ are open in $I$. This contradicts the connectedness of $I$, completing the proof that there is no clopen separation of $H_0$ and $H_1$ in $X$.
It follows immediately that $\beta X$ is not strongly zero-dimensional (Theorem 6.2.12 in Engelking) and since every Lindelöf zero-dimensional space is strongly zero-dimensional (this is easy to show), it follows that $\beta X$ is not even zero-dimensional.
Next, note that each $X_\eta$, being a subspace of a compact metrizable space $\big((\eta+1)\times I\big)$, is second countable, and therefore the weight of $X$ (= minimum cardinality of a base for the topology) is $\omega_1$. It follows that $X$ can be embedded in $D^{\omega_1}$, where $D=\{0,1\}$ with the discrete topology. Let $e:X\to D^{\omega_1}$ be such an embedding, and let $K=\operatorname{cl}e[X]$; $K$ is a zero-dimensional compactification of $X$ of weight at most $\omega_1$. Let $K^*=K\setminus e[X]$, and let $Z=\Big((\omega_2+1)\times K\Big)\setminus\Big(\{\omega_2\}\times K^*\Big)\;.$
Clearly $\{\omega_2\}\times e[X]$ of $Z$ is a closed subspace of $Z$ homeomorphic to $X$, so $\beta Z$ contains a closed subspace homeomorphic to $\beta X$, and we’ll have the desired example if we can show that $\beta Z$ is zero-dimensional. I’ll show that in fact $\beta Z=(\omega_2+1)\times K$ which is certainly zero-dimensional.
To do this, it suffices (e.g., by Corollary 3.6.4 in Engelking) to show that disjoint closed subsets of $Z$ have disjoint closures in $(\omega_2+1)\times K$. Suppose, then, that $F$ and $G$ are closed subsets of $Z$, let $\overline F$ and $\overline G$ be their closures in $(\omega_2+1)\times K$, and suppose that $p\in \overline F\cap\overline G$. Clearly $p\in \{\omega_2\}\times K^*$, so $p=\langle\omega_2,q\rangle$ for some $q\in K^*$. Let $\mathscr{B}_q$ be a local base of clopen sets at $q$ in $K$; since the weight of $K$ is at most $\omega_1$, we may assume that $|\mathscr{B}_q|\le\omega_1$ and index it as $\mathscr{B}_q=\{B_\xi:\xi<\omega_1\}$.
Enumerate $\omega_1\times\omega_1\times 2=\{\langle \alpha_\xi,\beta_\xi,i_\xi\rangle:\xi<\omega_1\}$. For each $\langle\eta,\xi\rangle\in \omega_2\times\omega_1$ let $V(\eta,\xi)=(\eta,\omega_2]\times B_\xi$, and let $\mathscr{V}=\big\{V(\eta,\xi):\langle\eta,\xi\rangle\in\omega_2\times\omega_1\big\}\;;$ $\mathscr{V}$ is a local base at $p$ in $(\omega_2+1)\times K$, so $F\cap V(\eta,\xi)\ne\varnothing\ne G\cap V(\eta,\xi)$ for each $\langle\eta,\xi\rangle\in\omega_2 \times\omega_1$. Thus, we can recursively construct a transfinite sequence
$\Big\langle\langle\eta_\xi,q_\xi\rangle:\xi<\omega_1\Big\rangle\tag{1}$
in $\omega_2\times K$ such that $\eta_\xi<\eta_{\xi+1}$ and $q_\xi\in B_{\alpha_\xi}$ for each $\xi<\omega_1$, $\langle\eta_\xi,q_\xi\rangle\in F$ whenever $i_\xi=0$, and $\langle\eta_\xi,q_\xi\rangle\in G$ whenever $i_\xi=1$.
Let $\eta=\sup\limits_{\xi<\omega_1}\eta_\xi<\omega_2$, and let p'=\langle\eta,q\rangle\in\omega_2\times K; the proof will be complete if I can show that p'\in F\cap G. To this end note that $\big\{(\xi,\eta]\times B_\alpha:\langle\xi,\alpha\rangle\in\eta\times\omega_1\big\}$ is a local base at p'. Fix $\alpha<\omega_1$; then
$\Big\langle\langle\eta_\xi,q_\xi\rangle:\alpha_\xi=\alpha\land i_\xi=0\Big\rangle$ and $\Big\langle\langle\eta_\xi,q_\xi\rangle:\alpha_\xi=\alpha\land i_\xi=1\Big\rangle$
are cofinal subsequences of $(1)$ lying entirely in $\big([0,\eta]\times B_\alpha\big)\cap F$ and $\big([0,\eta]\times B_\alpha\big)\cap G$, respectively, and it’s immediate that p'\in \operatorname{cl}_Z F=F and p'\in\operatorname{cl}_Z G=G, i.e., p'\in F\cap G. This completes the proof.