I tried Ex. 2.A.16 of Axler‘s "Linear Algebra Done Right".
Prob: Prove that the real vector space of all continuous real-valued functions on the interval $[0,1]$ is infinite-dimensional.
I found some solutions on the net and this one, on this page. Since my solution is different, I ask if my line of argumentation is correct.
Sol: The vector space of all real valued polynomials on the interval $[0,1]$ is an infinite-dimensional subspace of the real vector space of all continuous real-valued functions on the interval $[0,1]$, since polynomials are continuous on $[0,1]$. By contraposition of thm.2.26 (if a subspace $U$ of a vector space $V$ is infinite-dimensional, then $V$ is infinite-dimensional) the vector space of all continuous real-valued functions on the interval $[0,1]$ must be infinite-dimensional as well. $\blacksquare$
That $\mathcal{P}(\mathbb{R})$ is infinite-dimensional was proved in the text. I am accepting the reduction to the interval $[0,1]$ without proof. But such a proof would follow the same line as for all polynomials. Any comments and critique is welcome.
"comments and critique" :
(1) You should list out the Exact Statement of the theorem you want to use : "thm.2.26" is neither sufficient nor self-contained in the context of your MSE Post.
(2) You have to tweak 2.26 to limit it to $[0,1]$ which itself might be more unnecessary work.
(3) You have to ensure that 2.26 did not use what you are trying to prove here, other-wise it will be circular logic & hence invalid.
(4) It is not necessary to go to 2.26 involving Infinite-Dimensional Sub-Space to get the Contradiction.
We can just use $Dim(U) \le Dim(V)$ not involving infinity [[ https://proofwiki.org/wiki/Dimension_of_Proper_Subspace_is_Less_Than_its_Superspace ]]
That itself is enough to prove what you want , like this :
In Case the Continuous-real-functions $Dim(C)$ were not Infinite-Dimensional , then the Polynomial SubSpace $Dim(P)$ too must not be Infinite-Dimensional to let $Dim(C) \ge Dim(P)$
Since $Dim(P)$ is Infinite , $Dim(C)$ must also be Infinite
(5) Your Alternate thinking also gives ideas about various interconnections between various theorems.
(6) SUMMARY : Your line of argument is correct (over-all your Proof is nice too) but not complete. When you fill in the missing pieces & make simplifications , it might still be longer than what is given else-where like https://linearalgebras.com/2a.html