As said in the comments, the Lemma you mentioned cannot be used to establish the homogeneity of degree $0$ in prices and income of the demand functions, because it is only a necessary condition.
And the obviously straightforward way to show homogeneity is multiplying the variables in the budget constraint.
However, your question about whether it is possible to prove homogeneity by Euler's Theorem makes sense. Indeed, Euler's Theorem is a necessary and sufficient condition, and if demand is homogeneous it must hold, and the converse, if it holds the demand is homogeneous.
The question is how, practically, by calculations, to show this.
One has to proceed to find the derivatives of the demand function with respect to prices and income and verifying that they satisfy the Euler's Theorem.
How to compute these derivatives?
We can resort to the implicit function theorem, applying it to the first order conditions of the maximizing problem of the consumer, from which demand functions are derived.
I limit myself to the case of two goods, as for more goods the calculations are cumbersome, and also for two goods they require a bit of patience.
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Assuming the utility function differentiable as needed, and assuming local non satiation, so that the budget constraint is fulfilled as equality, the optimization problem is given by
$$u(x_1,x_2)\;\; \;\;\;max$$
$$ s. t. p_1x_1+p_2x_2=m$$,
where the variables have the usual meaning, $m$ is income.
Using Lagrange multipliers, $\lambda$, the first order conditions can be written as :
$$ m - p_1 x_1(p_1,p_2, m)- p_2 x_2(p_1,p_2, m) = 0. $$
$$ \frac{\partial U( x_1(p_1,p_2, m), x_2(p_1,p_2, m))}{\partial x_1} - \lambda p_1 = 0 $$
$$ \frac{\partial U( x_1(p_1,p_2, m), x_2(p_1,p_2, m))}{\partial x_2} - \lambda p_2 = 0 $$
The solutions of this system of equations are the demand functions $x_i(p_1,p_2, m)$, $i=1,2$.
Of course, it cannot be solved explicitly, but via the implicit function theorem (if its assumptions hold) the derivatives of the demand functions can be calculated.
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Let's calculate the derivatives of $x_1(p_1,p_2, m)$ with respect to $p_1, p_2, m$. This can made as follows$^1$.
Differentiating the first order conditions with respect to $p_1$ and writing it in matrix form we have:
$$\begin{bmatrix} 0 & -p_1 & -p_2 \\ -p_1 & u_{11} & u_{12} \\ -p_2 & u_{21} & u_{22} \end{bmatrix} \begin {bmatrix} \frac{\partial \lambda}{\partial p_1}\\\frac{\partial x_1}{\partial p_1}\\\frac{\partial x_2}{\partial p_1} \end{bmatrix}= \begin {bmatrix} x_1\\\lambda\\0\end {bmatrix}.\;\;\;\;\;\;\;\;\;\;(1)$$
Using Cramer's rule for systems we have:
$$\frac{\partial x_1}{\partial p_1}= \frac { \begin {vmatrix} 0 & x_1 & -p_2\\ -p_1& \lambda & u_{12}\\ -p_2 & 0& u_{22}\end {vmatrix} }{H} \;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2) $$
where $H\neq 0$ is the determinant of the bordered hessian (the determinant of the first matrix in $(1)$).
Expanding the determinant by the second column we have
$$\frac{\partial x_1}{\partial p_1}= \lambda \frac { \begin {vmatrix} 0 & -p_2 \\ -p_2 & u_{22} \end{vmatrix}}{H} -x_1 \frac {\begin {vmatrix} -p_1 & u_{12}\\-p_2 & u_{22}\end {vmatrix}} {H}. \;\;\;\;\;\;\;\;\;(3) $$
In the same way, differentiating the first order conditions with respect to $p_2$, and using Cramer’s rule, we obtain the derivative of $x_1$ with respect to $p_2$, that is
$$\frac {\partial x_1} {\partial p_2}=- \lambda \frac {\begin {vmatrix} 0&-p_2\\-p_1 & u_{12}\end {vmatrix}}{H} -x_2 \frac {\begin {vmatrix} -p_1& u_{12} \\-p_2 & u_{22}\end {vmatrix}}{H} \;\;\;\;\;(4) $$
The third derivative, differentiating the first order conditions with respect to $m$ , is
$$\frac {\partial x_1} {\partial m}= \frac {\begin {vmatrix} -p_1& u_{12}\\-p_2& u_{22}\end {vmatrix}}{H} \;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;(5)$$
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If you have the patience to make calculations (I consider a miracle I haven't mistaken some sign), you can verify that Euler's Theorem holds.
In fact, we have to verify that:
$$\frac {\partial x_1} {\partial p_1} p_1 +\frac {\partial x_1} {\partial p_2} p_2 +\frac {\partial x_1} {\partial m} m=0. \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(6)$$
Using expressions $(3$), $(4)$ and $(5$) for the derivatives, and neglecting $H$ we have:
$$(\lambda \begin {vmatrix} 0& -p_2\\-p_2& u_{22}\end {vmatrix} -x_1 \begin {vmatrix} -p_1& u_{12}\\-p_2& u_{22}\end {vmatrix}) p_1+ (-\lambda\begin {vmatrix} 0& -p_2\\-p_1& u_{12}\end {vmatrix}-x_2 \begin {vmatrix} -p_1& u_{12}\\-p_2& u_{22}\end {vmatrix}) p_2+ (\begin {vmatrix} -p_1& u_{12}\\-p_2& u_{22}\end {vmatrix}) m =0. \;\;\;\; (7)$$
Replacing $m$ with $p_1x_1+p_2x_2$ and making products we can verify that $(7)$, and therefore $(6)$, holds.
We have seen this way that Euler’s theorem holds for the demand of goods 1, $x_1(p_1, p_2, m)$
The same procedure can show that it holds for the demand of good 2, $x_2 (p_1,p_2, m)$.
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Let's say again this is not the most comfortable way to show that the demands are homogeneous of degree $0$. But it could be a useful exercise in calculating derivatives by the implicit functions theorem, and verifying the validity of Euler's Theorem in this case.
The use of the implicit functions theorem to calculate derivatives is important in economics, as it is a tool of comparative statics, both in microeconomics, as for demand functions, and in macroeconomics, for instance to calculate the multipliers.
$^1$$\scriptsize\text {When using 'differentials' in the implicit functions theorem, a warning must be given, to not convey a wrong message: the use}$
$\scriptsize\text {of 'differentials' as above is not rigorous, it is just a shortcut, a useful practical and mnemonic device to make calculations. But}$ $\scriptsize\text {if one looks at the implicit function theorem and its proof, there aren't differentials.}$
