Showing that the derivative of a vector-valued function is the the derivative of it's components using linear operators Recently in class we had shown the following to be true using a proof that's equivalent to this ProofWiki article:

Let $\vec{r}(t) = x(t)\vec{i} + y(t)\vec{j} + z(t)\vec{k}$ be a
vector-valued function on $\left(a,b\right) \rightarrow \mathbb{V}^3$
whose components $x(t), y(t), z(t)$ are differentiable real functions.
Then $\vec{r}$ is differentiable and:
$$\vec{r}'(t) = x'(t)\vec{i} + y'(t)\vec{j} + z'(t)\vec{k}$$

Now while the linked proof is straightforward enough, I was thinking about differentiation in the context of a linear operator. My argument was: since differentiable real functions form a vector space and differentiation was a linear operator, then the theorem above immediately follows from the linearity property. Like this:
Let $x(t), y(t), z(t)$ be vectors from the vector space of real differentiable functions and let
$$\vec{r}(t) = x(t)\vec{i} + y(t)\vec{j} + z(t)\vec{k}$$
Apply linear operator of differenitation and use linearity:
$$\vec{r}'(t) = \left(x(t)\vec{i} + y(t)\vec{j} + z(t)\vec{k}\right)' = x'(t)\vec{i} + y'(t)\vec{j} + z'(t)\vec{k}$$
When I asked my professor about this, he said that this cannot be done because I've applied linearity incorrectly, but upon further questioning he was unable to answer why. It seems valid to me, so I want to ask, is this a valid way to show the theorem? It seems much more elegant than using parametric limits.
The only possible problem I can see is the multiplication of vectors $x(t), y(t), z(t)$ with $\vec{i}, \vec{j}, \vec{k}$ since they are from different vector spaces. But since $x,y,z$ are real valued, it is, in a way, equivalent to scalar-vector multiplication.
 A: There are two claims you need to prove in order for your argument to be fully rigorous. $\newcommand{\r}{\mathbf r}$
Let $U$ be the vector space of $\mathbb R^3$-valued functions of one variable, each of whose components with respect to the standard basis vectors is differentiable, and $V$ the vector space of $\mathbb R^3$-valued functions of one variable.
Claim 1: Any $\r\in U$ is differentiable.
Claim 2: The map $T:U\to V$ defined by $T(\r)(t) = \r'(t)$ is linear.
Corollary: If $\r(t) = x(t)\mathbf i + y(t)\mathbf j + z(t)\mathbf k$, and each of $x,y,z$ are differentiable, then $\r$ is differentiable, and $\r'(t) = x'(t)\mathbf i + y'(t)\mathbf j + z'(t)\mathbf k$.
(The proof of this corollary is essentially what you gave in your post, minus the proof of differentiability of $\r$.)
Both claims technically require their own rigorous proofs, since we are not in the world of single-variable analysis, so the "constants" $\mathbf i,\mathbf j,\mathbf k$ and vector/scalar algebra do not a priori behave nicely with differentiation. This is not to say that giving such proofs is particularly difficult.
