Two definitions of complete algebraic reducibility are equivalent by Zorn's lemma. Does this equivalence stay for topological spaces? Consider a representation $A$ of a group $G$ in a complex vector space ${\mathbb{V}}\,$:
$$
 A:\;\;G\,\longrightarrow\,GL({\mathbb{V}})\;\;,
 $$
and let ${\mathbb{V}}$ be decomposable into a finite or infinite direct sum
$$
 {\mathbb{V}}\,=\,\bigoplus_i {\mathbb{V}}_i
 $$
of invariant subspaces carrying irreducible subrepresentations of $A\,$:
$$
 A(G)\,{\mathbb{V}}_i\subseteq{\mathbb{V}}_i\;\;,\qquad A(G)\,=\,\bigoplus_i \underbrace{A_i(G)}_{  \stackrel{\;}{\textstyle{\rm{irreducibles}}}   }
 \;.
 $$
Complete (algebraic) reducibility can be defined in two ways.
DEFINITION 1.
A representation $A$ is said to be completely reducible if it splits into a (finite or infinite) direct sum of irreducibles.
DEFINITION 2.
A representation $A$ is said to be completely reducible if any of its subrepresentations has a complementary subrepresentation.
THEOREM: $~~~$DEF 1 $~~\Longrightarrow~$ DEF 2
PROOF:
Let ${\mathbb{W}}\subseteq{\mathbb{V}}$ be an invariant subspace carrying a subrepresentation $A_W$ of $A$.
Our plan comprises two items:
(1) to prove the existence of a maximal (under inclusion) invariant subspace ${\mathbb{W}}^{\,\prime}\subseteq{\mathbb{V}}$ such that ${\mathbb{W}}$ and
${\mathbb{W}}^{\,\prime}$ are in direct sum, i.e. that ${\mathbb{W}}\cap {\mathbb{W}}^{\,\prime}=\vec{ 0}\,$;
(2) to show that it is a full complement of ${\mathbb{W}}$ in ${\mathbb{V}}$, i.e. that ${\mathbb{W}}\oplus{\mathbb{W}}^{\,\prime}={\mathbb{V}}\,$;
Consider the set $\cal D$ of all invariant subspaces ${\mathbb{U}}$
such that ${\mathbb{U}}\cap{\mathbb{W}}=\vec{ 0}\,$:
$$
  {\cal D}\,=\,\left\{ {\mathbb{U}}:\;\;\;{\mathbb{U}}\subset{\mathbb{V}}\,,\;\;\;A(G)\,{\mathbb{U}}\subseteq{\mathbb{U}}\,,
  \;\;\;{\mathbb{U}}\cap{\mathbb{W}}=\vec{ 0}  \right\}\;.
  $$
The set $\cal D$
- is not empty, because it includes $\{\vec{0}\}\;$,
- is partially ordered by inclusion $\subseteq\;$,
- contains totally ordered subsets.
For each totally ordered subset ${\cal{J}}\subset{\cal{D}}$, the union
$$
  \tilde{\mathbb U}_{\cal J}\,=\,\bigcup_{U\in\cal J} {\mathbb{U}}
  $$
- is a linear subspace of $\mathbb V\,$,
- is invariant and, therefore, is home to a subrepresentation of $A\,$,
- is an upper bound of the totally ordered subset ${\cal{J}}\subset\cal D$,
- is residing in $\cal D\,$.
Since any totally ordered subset of $\cal D$ has an upper bound in $\cal D$, then $\cal D$ contains, by Zorn's lemma, a maximal element, i.e. such an invariant subspace ${\mathbb{W}}^{\,\prime}$ in direct sum with $\mathbb W$ that, for any invariant ${\mathbb U}\in\cal D$, the condition ${\mathbb{W}}^{\,\prime}\subset \mathbb U$ implies  ${\mathbb{W}}^{\,\prime}= \mathbb U$. So item (1) is proved.
The proof of item (2) is borrowed from page 15 in the book by C. Gruson and V. Serganova https://math.berkeley.edu/~serganov/math252/Bookrep.pdf
$\cal D$ is by definition  a set of all subspaces in direct sum with $\mathbb W$. Thence, for an arbitrary irreducible ${\mathbb{V}}_i\notin\cal D$, we have ${\mathbb V}_i\cap{\mathbb W}\neq\vec{0}$ and, consequently,
$$
  ({\mathbb{W}}^{\,\prime}\oplus {\mathbb{V}}_i  )\cap{\mathbb{W}} \neq \vec{0}
  $$
From this fact and from the irreducibility of ${\mathbb{V}}_i$, follows
$$
  {\mathbb{V}}_i\cap({\mathbb W}^{\,\prime} \oplus 
  {\mathbb W})\neq \vec 0\;\;\Longrightarrow\;\;
  {\mathbb{V}}_i\subset({\mathbb W}^{\,\prime} \oplus {\mathbb W})\;\;\Longrightarrow\;\;{\mathbb W}^{\,\prime} \oplus {\mathbb W}\,=\,{\mathbb V}\;.
  $$
It proves item (2). This item warrants that ${\mathbb{W}}^{\,\prime}$ carries a subrepresentation $ A_{W^{\prime}}$ complementary to $ A_W$, i.e. satisfying $A_W \oplus A_{W^{\prime}}=A\,$.
QED
$\phantom{~}$
Generally, topological subspaces are not complemented. Still, if in DEF 1 we postulate that $A$ splits into irreducible topological components, we probably may expect that DEF 2 fulfils for all topological subspaces. Hence my
QUESTION
Is the above proof applicable to topological spaces?
Say, will it be legitimate to substitute everywhere "subspace" with "closed subspace"?
 A: This is a humble (and not necessarily correct) attempt to answer my own question.
Suppose our definitions now read:
DEFINITION 1.
$A(G)$ is $\,$topologically$\,$ completely reducible if it splits into a (finite or infinite) direct sum of $\,$topological$\,$ irreducibles.
DEFINITION 2.
$A(G)$ is $\,$topologically$\,$ completely reducible if any of its $\,$topological$\,$ subrepresentations has a complementary $\,$topological$\,$ subrepresentation.
Before proving that $\;$DEF 1 $\Longrightarrow$ DEF 2$\,$, we recall that the definition of a $\,$topological$\,$ subrepresentation implies not only the invariance but also the closedness of the corresponding subspace.
Now, to make the algebraic proof topological, we must:

*

*switch from vector spaces to topological vector spaces (TVS),

*switch from subrepresentations to topological subrepresentations,

*switch from irreducibles to topological irreducibles.

Specifically, our subspace ${\mathbb{W}}\subseteq\mathbb V$ is now closed. Also, we should now prove that the complement ${\mathbb{W}}^{\,\prime}\subseteq\mathbb V$ is closed also. To this end, we now define $\cal D$ as the set of all closed invariant subspaces in direct sum with $\mathbb W$.
At this point, we encounter a difficulty in our application of Zorn's lemma. While the infinite union $\tilde{\mathbb{U}}_{\cal J}$ of the closed invariant subspaces in a chain is invariant, it is not obliged to be closed.
To circumvent this difficulty, one may introduce the upper bound as the $\,$closure$\,$ of the infinite union of the closed invariant subspaces. If this step is legitimate, then the proof stays. However, the step is legitimate iff the closure is (a) invariant and (b) in direct sum with $\mathbb W$.
There is no obvious reason for any of these two statements to be universally correct. (I have created a separate question on this.) It therefore seems to me that the proof breaks here, and the two Definitions may be topologically inequivalent.
Comments and criticisms would be most appreciated.
