Consider the notion of a tree defined in descriptive set theory (see Classical Descriptive Set Theory by Kechris) as follows.
A tree on a set $A$ is a subset $T\subset A^{<\Bbb{N}}$ closed under initial segments; i.e., if $t\in T$ and $s\subset t$, then $s\in T$. In particular, $\varnothing\in T$ if $T$ is nonempty. We call the elements of $T$ the nodes of $T$. An infinite branch of $T$ is a sequence $x\in A^{\Bbb{N}}$ such that $x|n\in T$, for all $n$. The body of $T$, written as [$T$], is the set of all infinite branches of $T$. A tree $T$ pruned if every $s\in T$ has a proper extension $t\supset s,t\in T$.
I believe that the height of a tree is at most $\aleph_0$. I wonder if this has been proved already.
Note that another sense of the word "tree" is "partial order in which every down-set is linearly ordered; under this definition, every linear ordering is a tree. Also, note that we should be describing height and related notions by ordinals, not cardinals, so "height $\omega$" is better than "height $\aleph_0$."
Also, I'm adopting the minor additional convention that trees are nonempty. This is easily removed if you wish by changing slight details.
A tree on a set $A$ - that is, a subset of $A^{<\omega}$ closed under initial segments - does in fact have height at most $\omega$, trivially. Here I'm assuming that by "height" you're referring to the supremum of the ordertypes of the chains. To prove this, note that if $S$ is a set of finite strings which is linearly ordered by extension then the map from $S$ to $\omega$ sending each element of $S$ to its length is order-preserving.
However, there is also the notion of rank. This may feel like height at first but it behaves very differently. Because rank is so important and behaves nontrivially (as opposed to height) in this context, I'm going to give a brief summary of it here.
The most obvious difference between rank and height is that rank applies only to well-founded trees (or, we say ill-founded trees all have rank $\infty$); remember that a tree $T\subseteq A^{<\omega}$ is well-founded iff it has no infinite path, that is iff there is no $f\in A^\omega$ such that $f\upharpoonright n\in T$ for all $n\in\omega$. Well-founded trees are exactly those which we can induct along, and the rank of a tree essentially measures how "long" that induction is.
The rank of a well-founded tree$^*$ on an arbitrary set $A$ is defined recursively: the rank of a well-founded tree $T$ is the supremum of the ranks of the immediate subtrees (that is, those of the form $\{\sigma: n^\smallfrown\sigma\in T\}$ for $n\in\omega$). Note that this tells us that the tree consisting of just the root has rank $0$, since $0$ is the supremum of the emptyset.
Calculating the rank of a well-founded tree can be quite difficult, but there is one important global fact which is easy to establish:
In particular, the supremum of the ranks of well-founded trees on $\omega$ is $\omega_1$.
One direction is immediate - every well-founded tree on $A$ has at most $\vert A\vert$-many nodes, and it's easy to show that the rank of a well-founded tree can't exceed its cardinality.
For the other direction, fix $A$; we'll show by induction that for each $\alpha$ of cardinality at most $\vert A\vert$, there is a well-founded tree $T_\alpha$ on $A$ of rank $\ge\alpha$.
We let $T_0$ consist of just the root, and for $\alpha=\beta+1$ we let $T_\alpha=\{\langle\rangle\}\cup\{\langle 0\rangle^\smallfrown \sigma: \sigma\in T_\beta\}$ (here "$0$" denotes some fixed element of $A$). The interesting step is when $\alpha$ is a limit. For $\alpha$ a limit ordinal of cardinality $\le\vert A\vert$, fix a surjection $f: A\rightarrow \alpha$ and let $$T_\alpha=\{\langle\rangle\}\cup\{\langle i\rangle^\smallfrown\sigma: i\in A, \sigma\in T_{f(i)}\}.$$ It's easy to check that the rank of $T_\alpha$ is at least $\alpha$ (in fact, exactly $\alpha$) for each $\alpha$ of cardinality $\le\vert A\vert$, so we're done.
The notion of rank is extremely important in descriptive set theory and logic in general. Meanwhile, note that well-founded trees, despite not having any infinite paths, can have height $\omega$: consider the tree $$\{\langle\rangle\}\cup \{\sigma: \vert\sigma\vert<\sigma(0)\}.$$
$^*$Note that rank actually makes sense in a much broader context than just downwards-closed nonempty sets of finite strings from some nonempty set. As long as $P$ is a partial order with no infinite ascending sequence, the definition of rank makes sense and behaves well. We can similarly define the rank of an arbitrary partial order with no infinite descending sequence - either from scratch or as the rank of its "opposite" order.