Any $T:\mathbb{R}^3\to\mathbb{R}^3$ decomposes $\mathbb{R}^3$ into two nontrivial invariant subspaces? Since the characteristic polynomial of $T$ is degree 3, it has a real root. 
Either $f_T(x)$ has one real root, or 3. If $f_T(x)$ has 3 real roots:


*

*Case 1) $$f_T(x) = (x-\lambda_1)(x-\lambda_2)(x-\lambda_3)$$
with distinct $\lambda_1,\lambda_2,\lambda_3$. This case is easy.

*Case 2)
$$f_T(x) = (x-\lambda_1)(x-\lambda_2)^2$$
with $\lambda_1\ne \lambda_2$, then $(x-\lambda_1)$ and $(x-\lambda_2)^2$ are coprime, and result follows. 

*Case 3) 
$$f_T(x) = (x-\lambda_1)^3$$
I suspect there is counterexample here. Consider any Jordan block, somehow show that it cannot be transformed into two diagonal blocks, one of size 1, the other size 2.
If $f_T(x)$ has one real root, then 
$$f_T(x) = (x-\lambda)(x^2+bx+c)$$
again $(x-\lambda)$ and $(x^2+bx+c)$ are coprime and result follows. 
So the only case I don't know is where $f_T(x) = (x-\lambda)^3$. And I suspect that it is not possible to decompose $\mathbb{R}^3$ into two non-trivial invariant subspaces.
 A: Since as you said the characteristic polynomial $\chi_T$ has the degree $3$ then it has a real root $\lambda$ and then
$$\chi_T(x)=(x-\lambda)(x^2+ax+b)$$
In the case where the quadratic polynomial has  a real root the result is trivial. Assume now that the quadratic polynomial has complex non real roots so $\lambda$ isn't one of their roots and then
$$\gcd(x-\lambda, x^2+ax+b)=1$$
so by the lemma kernels$^{(1)}$ we have
$$V=\ker(T-\lambda\operatorname{id})\oplus \ker(T^2+aT+b\operatorname{id})$$ 
so $T$ admit a $T$-invariant two dimensional complement.
Edit Looking at the other answers,  the lemma mentioned in my answer is known as primary decomposition theorem.

$(1)$ This is a classic result in linear algebra, however I didn't find the Wikipedia page in English.
A: Your suspicion at the end of the question is correct, and Najib Idrissi's comment (which I've upvoted) is the most natural way to prove it. Unfortunately, his little comment is in danger of not being noticed in the presence of two wrong answers.  So this answer is just to make Najib's solution visible. The matrix
$$
\begin{pmatrix}0&1&0\\0&0&1\\0&0&0
\end{pmatrix}
$$ 
has only two non-trivial invariant subspaces, and their sum is not the whole space (in fact, one of the two is included in the other).
A: The characteristic polynomial of $T$ is the form  $(X-\lambda)Q(X)$ where $\deg Q=2$ and $Q(\lambda)\neq 0$, $i.e$ $X-\lambda$ and $Q$ are co-prime, by the primary decomposition theorem we have $\Bbb R^3=\ker(T-\lambda I_3)\oplus \ker(Q(T))$ and the subspace $\ker(Q(T))$ is $T$-invariant.  
