Closure properties can be formulated in terms of concepts from universal algebra. Let $X$ be the underlying set (in our examples, $X$ is a famly of sets itself). Let $I$ be an index set, $(\kappa_i)_{i\in I}$ be a family of cardinal numbers and $(f_i)_{i\in I}$ a family of function satisfying $f_i:X^{\kappa_i}\to X$ for all $i$. We say that $C\subseteq X$ is closed under $(f_i)_{i\in I}$ if we have for all $i\in I$ that $f_i(x)\in C$ for all $x\in C^{\kappa_i}$. One can show that the family of sets closed under $(f_i)_{i\in I}$ forms a Moore collection.
Let's look an an example: Let $U$ be a set and $X\subseteq 2^U$. We let $I=\{s,c,u\}$, $\kappa_s=0$, $\kappa_c=1$, and $\kappa_u=\omega$. We identify constants and nullary functions, so we can let $f_s=U$. We let $f_c(A)=A^C$ for all $A\in X$, and we let $f_u(A_0,A_1,\ldots)=\bigcup_n A_n$. That $X$ is closed under these three functions means simply that it contains $X$, is closed under complements and countable unions- it is a $\sigma$-algebra.
Now, one cannot write down semi-algebras this way, since there is no unique decomposition of the complement into disjoint sets. If $\mathcal{S}$ is a semi-algebra and $A\in\mathcal{S}$, then there exists a number $n$ and sets $B_1,\ldots,B_n\in\mathcal{S}$ that are disjoint and such that $A_c=B_1\cup\ldots\cup B_n$. Now if there exists a unique such family and if this family only depended on $A$, we could write down this property as closure under some functions in the following way: We let $f_{c_1}=B_1,\ldots, f_{c_n}=B_n$, and for $m>n$ we let $f_{c_m}=f_{c_n}$. We use the last condition because we have no a priori bound on how many sets are needed. But these sets are not a function of $A$, so this property can not be viewed as a closure property.
Here is an explicit example (taken from Alprantis & Border) that shows that the intersection of sem-algebras might fail to be a semi-algebra: Let $X=\{0,1,2\}$, $\mathcal{S}_1=\big\{\emptyset, X,\{0\},\{1\},\{2\}\big\}$, $\mathcal{S}_2=\big\{\emptyset, X,\{0\},\{1,2\}\big\}$, and $A=\{0\}$. We have $\mathcal{S}_1\cap\mathcal{S}_2=\big\{X,\emptyset,\{0\}\big\}$, and $A^C=\{0\}^C=\{1,2\}$ is not the disjoint union of elements of this intersection.
Following the tips of GEdgar i've realized that the answer was pretty simple, sometimes we just need some encouragement ^_^
Algebraic closure implies closure under unions of chains
Let $X$ be a chain with $X_1 \subseteq X_2 \subseteq ...$ (using countable index for simplicity).
$C(\bigcup X)=\bigcup\{C(Y)\ |\ \ (Y \subseteq \bigcup X) \land (|Y| \in \mathbb{N})\}$ (for algebraicity)
For every $Y$ we can find an $X_n$ such that $Y \subseteq X_n$, this prove $C(Y) \subseteq C(X_n)=X_n$. Taking the unions from both sides we get: $C(\bigcup X) \subseteq \bigcup X$
The other verse of inclusion is trivial, hence equality is established and $K$ is closed under unions of chains.
Closure under unions of chains implies algebraic closure
$C(X)=C(\bigcup\{Y\ |\ \ (Y \subseteq X) \land (|Y| \in \mathbb{N})\})=C(\bigcup\{C(Y)\ |\ \ (Y \subseteq X) \land (|Y| \in \mathbb{N})\})$
$\{C(Y)\ |\ \ (Y \subseteq X) \land (|Y| \in \mathbb{N})\}$ is upward directed so its union is closed and the algebraic proprerty is proved.
Best Answer
P.S. Here is a somewhat more recent reference which is freely available: George Markowsky, Chain-complete posets and directed sets with applications, Algebra Univ. 6 (1976), 53–68 (pdf).
That result was published by J. Mayer-Kalkschmidt and E. Steiner, Some theorems in set theory and applications in the ideal theory of partially ordered sets, Duke Math. J. 2 (1964), 287–289.
Inasmuch as I don't have access to the paper, which is hidden behind a paywall (the result is stated on the first page, which is freely available), the proof below may not be exactly the same as the one in the paper. I write "directed" for "directed upwards".
Lemma. If $D$ is a directed set of cardinality $\kappa$, an infinite cardinal, then $D$ is the union of a chain of directed subsets, each of cardinality less than $\kappa$.
Proof. We consider three cases.
Case 1. $\kappa=\aleph_0$.
Let $D=\{d_n:n\lt\omega\}$ be an enumeration of $D$. We will define a sequence of finite directed sets $D_n$. Let $D_0=\{d_0\}$. For $n\gt0$, having defined $D_{n-1}$, let $u_n$ be an upper bound for $D_{n-1}\cup\{d_n\}$, and let $D_n=D_{n-1}\cup\{d_n,u_n\}$. Then $\{D_n:n\lt\omega\}$ is a chain of finite directed sets whose union is $D$.
Case 2. $\kappa$ is an uncountable regular cardinal.
Let $D=\{d_\alpha:\alpha\in\kappa\}$. For $\beta\in\kappa$, let $D_\beta=\{d_\alpha:\alpha\lt\beta\}$. Then $B=\{\beta\in\kappa:D_\beta\text{ is directed}\}$ is unbounded in $\kappa$, so that $\{D_\beta:\beta\in B\}$ is a chain of directed sets whose union is $D$, and of course $|D_\beta|\lt\kappa$ for each $\beta$. (To see that $B$ is unbounded, given an ordinal $\beta_o\lt\kappa$, consider the limit of a sequence $\beta_0\lt\beta_1\lt\beta_2\lt\cdots\lt\beta_n\lt\cdots\lt\kappa$ where every finite subset of $D_{\beta_n}$ has an upper bound in $D_{\beta_{n+1}}$.)
Case 3. $\kappa$ is a singular cardinal.
Let $D=\bigcup_{\alpha\in\lambda}E_\alpha$ where $|E_\alpha|\lt\kappa$ and $\lambda=\operatorname{cf}\kappa$. Recursively define directed sets $D_\alpha\subseteq D\ (\alpha\in\lambda)$ so that $|D_\alpha|\lt\kappa$ and $E_\alpha\cup\bigcup_{\xi\lt\alpha}D_\xi\subseteq D_\alpha$. Then $\{D_\alpha:\alpha\in\lambda\}$ is a chain of directed sets whose union is $D$.
Theorem. If a family $K$ of sets is closed under unions of chains, then $K$ is closed under directed unions.
Proof. Assuming the contrary, let $D$ be a directed subfamily of $K$ of minimum cardinality whose union does not belong to $K$. Of course $D$ must be infinite. By the lemma, we can write $D=\bigcup_{\alpha\in\lambda}D_\alpha$, where $\{D_\alpha:\alpha\in\lambda\}$ is a chain of directed families, and $|D_\alpha|\lt|D|$ for each $\alpha$. By the minimality of $|D|$, we have $d_\alpha=\bigcup D_\alpha\in K$ for each $\alpha$. But then, since $\{d_\alpha:\alpha\in\lambda\}$ is a chain in $K$, we have $\bigcup D=\bigcup_{\alpha\in\lambda}d_\alpha\in K$, contradicting our assumption that $\bigcup D\notin K$.