[proofplan] The proof is a direct verification of the definition of an isometric embedding. We first show that the inclusion preserves the distance between every pair of points, using exactly the defining formula for the subspace metric. We then verify injectivity of the inclusion map, which completes the definition of an isometric embedding. [/proofplan]

[step:Verify that the inclusion preserves all distances]Let $y_1,y_2\in Y$. Since $i:Y\to X$ is the inclusion map, $i(y_1)=y_1$ and $i(y_2)=y_2$ as points of $X$. By the definition of the subspace metric $d_Y$ on $Y$, \begin{align*} d_Y(y_1,y_2)=d_X(y_1,y_2). \end{align*} Therefore, \begin{align*} d_X(i(y_1),i(y_2))=d_X(y_1,y_2)=d_Y(y_1,y_2). \end{align*} Thus $i$ preserves distances between all pairs of points in $Y$.[/step]

[guided]We need to prove that the map $i:Y\to X$ is distance-preserving. This means that for every pair of points $y_1,y_2\in Y$, the distance between their images in $X$ must equal their distance inside $Y$. Let $y_1,y_2\in Y$. The inclusion map does not change the underlying point: by definition, $i(y_1)=y_1$ and $i(y_2)=y_2$. The only possible issue is that the same two points are being measured with two different metrics: $d_Y$ on the domain $Y$, and $d_X$ on the ambient space $X$. But $d_Y$ is the subspace metric, so it is defined by restricting $d_X$ to pairs of points in $Y$. Hence \begin{align*} d_Y(y_1,y_2)=d_X(y_1,y_2). \end{align*} Substituting $i(y_1)=y_1$ and $i(y_2)=y_2$ into the ambient distance gives \begin{align*} d_X(i(y_1),i(y_2))=d_X(y_1,y_2)=d_Y(y_1,y_2). \end{align*} Since $y_1$ and $y_2$ were arbitrary points of $Y$, the inclusion map preserves every distance.[/guided]

[step:Verify that the inclusion is injective] Let $y_1,y_2\in Y$ and suppose $i(y_1)=i(y_2)$. Since $i(y)=y$ for every $y\in Y$, this equality is exactly $y_1=y_2$. Hence $i$ is injective. [/step]

[step:Conclude that the inclusion is an isometric embedding] The previous steps show that $i:Y\to X$ is injective and satisfies \begin{align*} d_X(i(y_1),i(y_2))=d_Y(y_1,y_2) \end{align*} for all $y_1,y_2\in Y$. Therefore, by the definition of an isometric embedding, $i$ is an isometric embedding. [/step]