Let $A,B$ be matrices associated to $U,T$. You could look for two matrices such that $AB=0$, $BA\ne AB$, e.g. you could look for two diagonalizable, but not simultaneously diagonalizable, matrices (see Simultaneous diagonalization of commuting linear transformations).
Example: $U(x,y)=(-2x+y,-2x+y)$, $T=(x+2y,2x+4y)$, i.e.
$$A=\begin{bmatrix} -2 & 1 \\ -2 & 1 \end{bmatrix},\quad
B=\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$$
and $AB=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$, but $BA=\begin{bmatrix} -6 & 3 \\ -12 & 6 \end{bmatrix}$. So $UT(x,y)=AB(x,y)=(0,0)$, but $TU(x,y)=BA(x,y)=(-6x+3y, -12x+6y)\ne (0,0)$ for all $x\ne y/2$.
There's a bit of a logical mixup here.
You want to show, $x\mapsto\sqrt{\langle x,x\rangle}$ is a norm on any inner product space). The only axiom of a norm that is not immediate is the triangle inequality.
So you want to bound: $$\sqrt{\langle x+y,x+y\rangle}\le\sqrt{\langle x,x\rangle+\langle y,y\rangle+2\langle x,y\rangle}\le\sqrt{\langle x,x\rangle+\langle y,y\rangle+2\sqrt{\langle x,x\rangle\cdot\langle y,y\rangle}}$$Using the CBS inequality, and then from there you just observe: $$\sqrt{\langle x,x\rangle+\langle y,y\rangle+2\sqrt{\langle x,x\rangle\cdot\langle y,y\rangle}}=\sqrt{\langle x,x\rangle}+\sqrt{\langle y,y\rangle}$$Concluding the triangle inequality.
Now you object, because the CBS inequality is often stated as: $$\langle a,b\rangle^2\le\|a\|^2\|b\|^2$$Where $\|\cdot\|$ is "already a norm", or something. But really, when stating the CBS inequality and when proving it, the symbol "$\|\cdot\|$" is simply a placeholder for the symbol $\sqrt{\langle,\rangle}$ and the fact that $\|\cdot\|$ - so defined - is a genuine norm is not used.
So there is no problem at all.
The standard proof of the CBS inequality (real case): (this aspect of the post is very much a duplicate, save for the avoidance of norm notation - see here and here)
Fix $a,b$ and consider the map: $$f:\Bbb R\ni t\mapsto\langle a+tb,a+tb\rangle\in\Bbb R$$
We have: $$f(t)=t^2\langle b,b\rangle+2t\langle a,b\rangle+\langle a,a\rangle$$But also: $$f\ge0$$
So the discriminant must be nonpositive, i.e.: $$(2\langle a,b\rangle)^2-4\langle a,a\rangle\cdot\langle b,b\rangle\le0$$And then: $$\langle a,b\rangle^2\le\langle a,a\rangle\cdot\langle b,b\rangle$$Follows. Then: $$|\langle a,b\rangle|\le\sqrt{\langle a,a\rangle\cdot\langle b,b\rangle}$$Follows. This inequality can be used in the above proof that $x\mapsto\sqrt{\langle x,x\rangle}$ is a norm. There is absolutely no problem, and nowhere was the $\|\cdot\|$ notation needed or used.
If the space is a complex inner product space we use a very similar approach. Fix $a,b$ as before and define $f$ in the same way. Let $g:\Bbb C\to\Bbb R$ be the map: $$\lambda\mapsto|\lambda|^2\cdot\langle a,a\rangle+2|\lambda|\cdot|\langle a,b\rangle|+\langle b,b\rangle$$
Because: $$\begin{align}g(\lambda)&\ge|\lambda|^2\cdot\langle a,a\rangle+2\cdot\Re(\lambda\cdot\langle a,b\rangle)+\langle b,b\rangle\\&=\langle \lambda a+b,\lambda a+b\rangle^2\\&=f(|\lambda|)\\&\ge0\end{align}$$
For all $\lambda\in\Bbb C$, we conclude that the polynomial: $$\Bbb R\ni x\mapsto\langle a,a\rangle\cdot x^2+2|\langle a,b\rangle|\cdot x+\langle b,b\rangle\in\Bbb R$$Is nonnegative and so has nonpositive discriminant.
This discriminant is basically the same one as before, except with a modulus sign - you get: $$4|\langle a,b\rangle|^2-4\langle a,a\rangle\cdot\langle b,b\rangle\le0\implies|\langle a,b\rangle|^2\le\langle a,a\rangle\cdot\langle b,b\rangle$$
Best Answer
Yes, it correct! Well done! Basically this shows that every linear map between finite dimensional vector spaces is continuous.
One suggestion: I would write the basis $\mathcal{B}'= \{f_1, \dots, f_m\}$ because you already used $e_1, \dots, e_n$.
Maybe one nitpick: the question asks for $M > 0$. If $T= 0$, your proof gives $M=0$, but you can simply add $1$ to your choice of $M$.