There are several ways to explain "computable implies continuous", I shall give here two such explanations.
Turing machines compute continuous maps
Suppose we have a Turing machine which takes possibly infinite amount of input written down on an input tape. It writes the result on an output tape, where the output cells are write-once. There are working tapes. The machine may run forever, filling up the outputs cells. This is known as a type two machine. (The argument for other kinds of machines is going to be similar but simpler.)
The following should be obvious: when the machine writes into an output cell, its workings up to that point only depends on a finite portion of the input tape, for the simple reason that in finitely many computation steps it could not have moved the input head past some point. Therefore, every input tape that agrees with the given one up to that point would have caused the machine to write the same answer to the same output cell.
But this is a form of continuity, if we put the correct topology on the spaces of input and output tapes.
First we put topology on the set $\Sigma$ of symbols that can be written onto tape cells. For this we just pick the discrete topology. A tape is an infinite sequence of symbols, so an element of $\Sigma^\omega$, which is a product of $\Sigma$'s. Let's put the product topology on it.
Recall that a basic open set on $\Sigma^\omega$ for the product topology is of the form $U(a_0, \ldots, a_n) = \{\alpha \in \Sigma^\omega \mid \forall i < n . \, \alpha_i = a_i\}$, where $a_0, \ldots, a_n \in \Sigma$. That is, a basic open set fixes an initial portion of the sequence to the given values $a_0, \ldots, a_n$.
Now we may check that the function $f : \Sigma^\omega \to \Sigma^\omega$ computed by the machine is indeed continuous. Take a basic open set $V = U(a_0, \ldots, a_n)$ and let $W = f^{-1}(V)$. We need to verify that $W$ is open. For this purpose, consider any $\alpha \in W$. If we find a basic open set $W'$ such that $\alpha \in W' \subseteq W$, then we're done.
Because $\alpha \in W$, we have $f(\alpha) \in V$. Thus on input $\alpha$ the machine produces an output tape starting with $a_0, a_1, \ldots, a_n$. By the time it writes out these cells it has inspected at most the first $k$ cells of the input, for some $k \in \mathbb{N}$. We may take $W' := U(\alpha_0, \ldots, \alpha_k)$ and verify that $\alpha \in W' \subseteq W$. It is obvious that $\alpha \in W'$. To prove $W' \subseteq W$ take any $\beta \in W'$ and observe that $f(\alpha)$ and $f(\beta)$ agree on the first $n$ values of the output. This implies that $f(\beta) \in V$ and hence $\beta \in W$, as required.
Computable maps are continuous as maps between algebraic $\omega$CPOs
First, let me note that what you defined is usually called "$\omega$CPO" ($\omega$ in the name indicates we only require suprema of chains).
In denotational semantics datatypes correspond to $\omega$CPOs. In fact, they correspond to algebraic $\omega$CPOs (is this in your thesis?), which are $\omega$CPOs for which the compact elements form a base. Here are some definitions.
Definition: Let $D$ be an $\omega$CPO. An element $d \in D$ is compact if, for every $\omega$-chain $x_0 \leq x_1 \leq \cdots$ such that $d \leq \bigsqcup_i x_i$, there exists $j$ such that $d \leq x_j$.
Definition: An $\omega$CPO is algebraic if every $x \in D$ is the supremum of compact elements below it.
The intuition behind compact elements is that they contain "finite information".
A good example is $\mathcal{P}(\mathbb{N})$, the powerset of natural numbers ordered by $\subseteq$, where the compact elements are precisely the finite subsets of $\mathbb{N}$ (exercise!).
Another example: in the $\omega$CPO of continuous functions $\mathbb{N} \to \mathbb{N}_\bot$ the compact elements are those partial functions that are equal to $\bot$ everywhere, except at finitely many arguments.
To say that an $\omega$CPO is algbraic is to say that every element is completely determined by the finite pieces of information that approximate it. It is a fact that in denotational semantics datatypes correspond to algebraic $\omega$CPOs, unless we are doing something very unusual.
We can now explain why every computable map is continuous. Suppose $D$ and $E$ are $\omega$CPOS and $f : D \to E$ computable. Suppose $x \in D$, $e \in E$, $e$ is compact, and $e \leq f(x)$. Intuitively, this say that "finite piece of information $e$ appears in the output $f(x)$". Because $f$ is computable, it must be the case that it computed the information $e$ by accessing only a finite amount of information about $x$, i.e., there is a compact $d \in D$ such that $d \leq x$ and $e \leq f(d)$. This argument should be compared to the Turing machine argument above. We have established:
Lemma: If $f : D \to E$ is computable and $e \leq f(x)$ for some $x \in D$ and a compact $e \in E$, then there is compact $d \in D$ such that $d \leq x$ and $e \leq f(d)$.
We can usethe lemma to show that a computable $f$ is continuous. Suppose $x_0 \leq x_1 \leq \cdots$ is a chain in $D$. Because $f$ is monotone, we already know that $\bigsqcup_i f(x_i) \leq f(\bigsqcup_i x_i)$, but we also need the inequality $f(\bigsqcup_i x_i) \leq \bigsqcup_i f(x_i)$. Because $E$ is algebraic, it suffices to show that, whenever $e \in E$ is compact and $e \leq f(\bigsqcup_i x_i)$ then $e \leq \bigsqcup_i f(x_i)$. So assume $e \leq f(\bigsqcup_i x_i)$. By the lemma there exists a compact $d \in D$ such that $d \leq \bigsqcup_i x_i$ and $e \leq f(d)$. Because $d$ is compact, there exists $j$ such that $d \leq x_j$, hence by monotonicity of $f$ we have $e \leq f(d) \leq f(x_j) \leq \bigsqcup_i f(x_i)$. We are done.
