No. Indeed $\mathrm{SO}(4)$ satisfies the following condition (which is a first-order existential formula) but not $\mathrm{SU}(3)$:

$$\exists w,x,y,z: [x,w]\neq 1\neq [y,z],\; [w,y]=[w,z]=[x,y]=[x,z]$$

(it just says there are two commuting pairs of non-abelian subgroups).

Indeed $\mathrm{SO}(4)$ even contains a copy of the direct product of two non-abelian free groups.

Let's show this doesn't exist in $\mathrm{U}(3)$. Suppose by contradiction we have such a 4-tuple. Since the $\mathbf{C}$-subalgebra generated by $w,x$ is semisimple and non-commutative, after conjugation it is either $\mathrm{M}_3(\mathbf{C})$ or the block-diagonal $\mathrm{M}_2(\mathbf{C})\times \mathrm{M}_1(\mathbf{C})$. So its centralizer is commutative in both cases, contradiction.

In particular there is no injective homomorphism $\mathrm{SO}(4)\to\mathrm{U}(3)$ (no continuity required).

Here's an alternative argument to show that the centralizer of every non-solvable subgroup

$G$ in

$\mathrm{GL}_3(C)$, with

$C$ any field, is abelian. We can suppose that

$C$ is algebraically closed. Indeed, choose a triangulation by blocks of

$G$ with irreducible diagonal blocks. If it's trivial (i.e.,

$G$ is irreducible), the centralizer is reduced to scalars. So it's

$1+2$ or

$2+1$ (if it were

$1+1+1$,

$G$ would be solvable). In both case, the subalgebra generated by

$G$ contains a diagonal copy of

$\mathrm{M}_2(C)\oplus \mathrm{M}_1(C)$. So its centralizer is commutative.

Since $\mathrm{SO}(4)$ contains two pairwise copies of the binary icosahedral group of order $120$, it also shows that some finite subgroup of $\mathrm{SO}(4)$ does not embed into $\mathrm{SU}(3)$ (and even in $\mathrm{GL}_3$).

Also, as pointed out by Friedrich Knop in a comment, the premise of the question is slightly false: $\mathrm{SO}(5)$ actually has no injective homomorphism into $\mathrm{SU}(4)$, regardless of any continuity assumption. Indeed, $\mathrm{SO}(5)$ has a 2-elementary subgroup of order $2^4$, while $\mathrm{SU}(4)$ has none. (One indeed has a continuous homomorphism $\mathrm{Spin}(5)\to\mathrm{SU}(4)$, or $\mathrm{SO}(5)\to\mathrm{PSU}(4)$.) So in this case the question is sensitive to coverings.

In contrast, no connected Lie group locally isomorphic to $\mathrm{SO}(4)$ (these are $\mathrm{SU}(2)^2$, $\mathrm{SO}(3)^2$, $\mathrm{SO}(3)\times\mathrm{SU}(2)$, and $\mathrm{SO}(4)$) can be embedded abstractly into a connected Lie group locally isomorphic to $\mathrm{SU}(3)$ (these are $\mathrm{SU}(3)$ and $\mathrm{PSU}(3)$). In each case some finite subgroup (which can be chosen as central quotient of the direct square of the binary icosahedral group of order $120$) testifies the obstruction.

isomorphicto a subgroup. $\endgroup$