Prove for a monotone, continuous, and rational preference relation $\succsim$ on $X=\mathbb{R}^L_+$, $y\geq x$ implies $y\succsim x$.

I need to prove the following result:

For a monotone, continuous, complete, and transitive preference relation $$\succsim$$ on $$X=\mathbb{R}^L_+$$, $$y\geq x$$ implies $$y\succsim x$$.

I tried it myself, but I'm not sure whether it is correct. May I ask someone to check my work? Thank you so much!

Let $$\succsim$$ be a monotone, continuous, complete, and transitive preference relation on $$X=\mathbb{R}^L_+$$. Suppose first that $$y\gg x$$; that is, $$y_i>x_i$$ for all $$i=1,\dots,L$$. Then by the definition of monotonicity, $$y\succ x$$ and thus $$y\succsim x$$.

Now suppose that $$y_i=x_i$$ for some $$i\in\{1,\dots,L\}$$ and $$y_k>x_k$$ for all $$k\neq i$$. (So $$y\geq x$$.) Let $$\{y^n\}$$ be a sequence of points in $$\mathbb{R}^L_+$$ such that $$y_k^n=y_k$$ for $$k\neq i$$ and $$y_i^n=y_i+\frac{1}{n}$$. Let $$\{x^n\}$$ be a sequence of points in $$\mathbb{R}^L_+$$ such that $$x_k^n=x_k$$ for $$k\neq i$$ and $$x^n_i=x_i+\frac{1}{n+1}$$. Then for each $$n$$, $$y^n_k = y_k > x_k = x^n_k$$ and $$y^n_i = y_i + \frac{1}{n} = x_i + \frac{1}{n} > x_i + \frac{1}{n+1} = x^n_i$$. Thus $$y^n\gg x^n$$, and monotonicity implies $$y^n\succ x^n$$, and so $$y^n\succsim x^n$$. So we have constructed a sequence of pairs $$\{(x^n,y^n)\}$$ such that $$y^n\succsim x^n$$ for all $$n$$, $$x=\lim_{n\to\infty}x^n$$, and $$y=\lim_{n\to\infty}y^n$$. Then continuity implies $$y\succsim x$$.

Some Background Information:

Notation$$\quad$$ $$\mathbb{R}^L_+ = \left\{x\in\mathbb{R}^L:x_l\geq0\ \text{for}\ l=1,\dots,L\right\}$$.

Notation$$\quad$$ Let $$x,y\in\mathbb{R}^N$$ so $$x=(x_1,\dots,x_N)$$ and $$y=(y_1,\dots,y_N)$$. Then

(i) $$x\geq y$$ means $$x_n\geq y_n$$ for all $$n=1,\dots,N$$;

(ii) $$x\gg y$$ means $$x_n>y_n$$ for all $$n=1,\dots,N$$.

Definition$$\quad$$ The preference relation $$\succsim$$ on $$X$$ is rational if it possesses the following two properties:

(i) Completeness. For all $$x,y\in X$$, we have $$x\succsim y$$ or $$y\succsim x$$ (or both).

(ii) Transitivity. For all $$x,y,z\in X$$, if $$x\succsim y$$ and $$y\succsim z$$, then $$x\succsim z$$.

Definition$$\quad$$ The preference relation $$\succsim$$ on $$X$$ is monotone if $$x\in X$$ and $$y\gg x$$ implies $$y\succ x$$.

Definition$$\quad$$ The preference relation $$\succsim$$ on $$X$$ is continuous if it is preserved under limits. That is, for any sequence of pairs $$\{(x^n,y^n)\}_{n=1}^{\infty}$$ with $$x^n\succsim y^n$$ for all $$n$$, $$x=\lim_{n\to\infty}x^n$$, and $$y\lim_{n\to\infty}y^n$$, we have $$x\succsim y$$.