Is the following Proof Correct?
Theorem. Given that $R$ is a total-order on $A$, every finite, non-empty set $B\subseteq A$ has an $R-$smallest element.
Proof. We construct the proof by recourse to Mathematical-Induction.
Basis-Step: Let $B$ be any non-empty subset of $A$ such that $|B|=1$ evidently the one element in $B$ is its $R-$smallest element since $R$ is reflexive.
Inductive-Step: Given any arbitrary $k\in\mathbf{Z^+}$ assume that the following is true $$\forall X\subseteq A(|X|=k\implies\exists a\in X\forall x\in X(aRx))\tag{1}$$ Now let $Y$ be an arbitrary \textit{non-empty} subset of $A$ such that $|Y|=k+1$. We could express $Y$ as $Y=X\cup\{\beta\}$ where $X=Y\backslash\{\beta\}$ and $\beta$ is an arbitrary element of $Y$, evidently $|X|=k$ and thus from $(1)$ it follows that $X$ has an $R-$smallest element.
Let this element by $\alpha$, moreover since $R$ is a total-order it follows that $$\forall x\in A\forall y\in A(xRy\lor yRx)\tag{2}$$ thus in particular we have $\alpha R\beta\lor\beta R\alpha$ arguing from cases.
Case-1($\alpha R\beta$): We know that $\forall x\in X(\alpha Rx)$ this together with $\alpha R\beta$ implies that $\forall x\in Y(\alpha Rx)$ consequently $\alpha$ is the $R-$smallest element of $Y$.
Case-2($\beta R\alpha$): Let $x\in X$ we know that $\alpha Rx$ this together with $\beta R\alpha$ and the transitivity of $R$ implies that $\beta Rx$ and the reflexivity of $R$ implies that $\beta R\beta$ consequently $\beta$ is the $R-$smallest element of $Y$.
$\blacksquare$