Let $U' = \text{Span}(u_1, \dots, u_{m-1})$, and $W = \text{Span}(u_m). \ $ We know that $V \cap U' = 0$ or else there would be $t$ in both spaces such that $t = \sum a_i v_i = \sum b_i u_i$, but by linear indep. we must have all $a_i, b_i = 0$.
Clearly $U = U' + W$. Suppose that there is a $v \in V \cap U$, i.e. by the that decomposition, $v = u' + w$, where $u' \in U', w \in W. \ $ But then $u' = v - w \in V$, since $u_m \in V$ is given, but since $u' \in U'$, by our first paragraph we have $u' = 0$. Thus $v = w \in W$. We have essentially shown that $V \cap U \subset W$. Conversely, if $v \in W$, then its easy to see that $v \in V \cap U$. Thus we've identified the intersection of the two spaces with $W = \text{Span}(u_m)$, which clearly has dimension $1$.
Technically, yes, it is a valid proof, if the vector spaces are assumed to be finite dimensional (which they don't seem to be), and the principle you quoted is correct.
However the results work for infinite dimensions too, and the dimension counts you made are no longer valid, so without the finite-dimensionality argument you need another proof.
Let me try to convince you, thoigh, that even if the vector spaces were assumed to be finite dimensional, you should do another proof.
Indeed, a very important first point is that this proof doesn't give you an explicit isomorphism, and in the examples you gave, there actually are very interesting explicit isomorphisms. They are interesting for a bunch of reasons :
First of all, they are explicit, and having explicit isomorphisms is always a good thing in practice. Here, one is often brought to identify, say $L(V\times W, E)$ with $L(V,E)\times L(W,E)$ and to make this identification you need the specific isomorphism.
Second of all, the explicit isomorphisms I am thinking of (which are almost surely the ones the author had in mind) are very nice, in that they're "natural". We could get into the technicalities to know what this means precisely, but essentially it means they behave well with the whole of linear algebra. By that I mean that for instance, regarding the last one, if you have two vector spaces $V,W$ and identify $V^n$ and $L(\mathbb F^n, V)$ via that isomorphism, and $W^n$ and $L(\mathbb F^n, W)$ via that isomorphism too, then nothing “goes wrong“ when you look at maps between $V$ and $W$. You can try to fiddle around with it a bit to see what I mean.
And lastly, note that although the course only mentions the principle for finite-dimensional spaces, the isomorphisms I have in mind work for infinite-dimensional spaces too, and the proof remains unchanged. That is often something that happens : doing proofs with dimensions is often more restrictive than doing "synthetic" proofs, and so the latter is better. Indeed, it often happens that there is extra structure on your spaces, and the dimension proofs don't see that at all, whereas proofs that actually exhibit the isomorphism are more likely to help you with that extra structure.
All of this is very common in linear algebra and other disciplines : students tend to use bases, dimensions, charts in differential geometry etc. whenever they have the opportunity, whereas it's often much better, for various reasons, to stay at the "synthetic" level, and the gain is often huge, although sometimes hard to see for students.
Best Answer
This is a common theorem that qualifies as
• Theorem 6.25 on P512 of Linear Algebra by David Poole,
• Theorem 3.18 on P55 of Linear Algebra Done Right by Sheldon Axler,
• and can be found here at Google Books.
Also, in your first line, I think that you should specify $\phi$ to be a bijective linear transformation, and not just a bijection. I hope that this helps.