I interpret the question as follows. Let $SU(3)$ be the group of complex $3\times 3$ matrices with $U^\dagger U=I$ and $\det U=1$ acting as linear operators in $\mathbb C^3$. Is there a subspace $M \subset \mathbb C^3$ with $M \neq \{0\}$, $M \neq \mathbb C^3$ such that $U(M) \subset M$ for every $U \in SU(3)$?
In other words, is the natural action of $SU(3)$ on $\mathbb C^3$ irreducible?
Your proof is wrong, because the fact that $SU(3)$ is not Abelian does not imply that it is irreducible.
Actually the irreducibility property is true and it holds for the natural action of $SU(N)$ in $\mathbb C^N$. I sketch a proof in the rest of my answer. Consider $x,y\in \mathbb C^N\setminus \{0\}$. It therefore holds $y = cx/||x|| + dy'$, where $y' \perp x$ and $||y'||=1$ and $c,d \in \mathbb C$.
Working in the plane generated by the orthonormal vectors $x/||x||$ and $y'$, it is easy to construct $U \in SU(N)$ such that $U x/||x|| = y/||y||$ using elementary computations (*).
Therefore $y= s Ux$ for some $s\in \mathbb C$ and $U \in SU(N)$.
We have established the following lemma.
If $x,y\in \mathbb C^N\setminus \{0\}$, there is $U \in SU(N)$ and $s \in \mathbb C$ such that $sUx= y$.
This fact implies that the action of $SU(N)$ on $\mathbb C^N$ is irreducible. Let us prove it. Suppose that $M \subset \mathbb C^N$ is a subspace invariant under $SU(N)$. If $M\ni x\neq 0$ and $y \in \mathbb C^N$, there is $U\in SU(N)$ such that $sUx =U(sx) =y$ for some constant $s$. As $sx \in M$ and $M$ is invariant, we have $U(sx) = y \in M$. Since $y$ is arbitrary, we finally obtain $M= \mathbb C^N$. This means that $SU(N)$ is irreducible on $\mathbb C^N$.
(*) $U$ can be costructed as follows. If $x/||x||$ and $y'$ are parallel everything is trivial. In the other case, $x/||x||, y', e_3,\ldots, e_N$ is an orthonormal basis of $\mathbb C^N$ for suitable orthogonal unit vectors $e_3,\ldots, e_N$. The unique linear map such that $$U : x/||x|| \mapsto \frac{cx/||x|| + dy'}{\sqrt{|c|^2+|d|^2}}$$ $$U: y' \mapsto ax/||x|| + by'$$
$$U: e_j \mapsto e_j $$
is unitary if and only if $$|a|^2+|b|^2 =1\tag{1}$$ and $$a\overline{c}+ b \overline{d}=0\:.\tag{2}$$ Here $d$ and $c$ are fixed and the system of equations (1)-(2) in $a, b \in \mathbb C$ always admit solutions.
Notice that if $(a,b)$ is a solution, $(\lambda a, \lambda b)$ is such, for $|\lambda|=1$.
Representing $U$ with respect to the said basis, since $U$ is represented by a unitary matrix, its determinant must be a unit complex. In other words $$|\det U| =|cb -da|/\sqrt{|c|^2+|d|^2}=1$$ so we can always fix the multiplier $\lambda$ in order to have $\det U=1$. However it does not automatically imply that $U \in SU(N)$, because the determinant is computed with respect to the wrong basis. Indeed, the relevant determinant used in the definition of $SU(N)$ is the one computed with respect to the canonical basis of $\mathbb C^N$. It is $${\det}_C U = \det M U M^{-1}$$ where $M$ is the unitary matrix relating the canonical basis of $\mathbb C^N$ to $x/||x||, y', e_3,\ldots, e_N$. Therefore $${\det}_C U = \det M \det U (\det M)^{-1} = \det U=1\:.$$ Summarizing, we have obtained that the found operator $U$, represented with respect to the canonical basis of $\mathbb C$ is a unitary matrix with determinant $1$. In other words $U$ belongs to $SU(N)$ as wanted.
