[proofplan]
We show that any convergent sequence in $Y$ has its limit in $Y$ by expressing convergence in terms of coordinates with respect to a fixed basis. The key ingredient is that all norms on a finite-dimensional space are equivalent, so coordinate convergence and norm convergence coincide.
[/proofplan]
[step:Fix a basis and express elements of $Y$ in coordinates]
Let $\dim Y = d$, and let $\{e_1, \ldots, e_d\}$ be a basis for $Y$. Every element $y \in Y$ has a unique representation
\begin{align*}
y = \sum_{i=1}^{d} \alpha_i \, e_i, \quad \alpha_i \in \mathbb{K},
\end{align*}
where $\mathbb{K}$ is the scalar field ($\mathbb{R}$ or $\mathbb{C}$). Define the coordinate map
\begin{align*}
T: \mathbb{K}^d &\to Y \subset X \\
(\alpha_1, \ldots, \alpha_d) &\mapsto \sum_{i=1}^{d} \alpha_i \, e_i.
\end{align*}
This map is a linear bijection from $(\mathbb{K}^d, \|\cdot\|_1)$ onto $(Y, \|\cdot\|_X)$, where $\|(\alpha_1, \ldots, \alpha_d)\|_1 := \sum_{i=1}^{d} |\alpha_i|$.
[/step]
[step:Establish equivalence of the coordinate norm and the $X$-norm on $Y$]
The map $T$ is continuous: for $\alpha = (\alpha_1, \ldots, \alpha_d) \in \mathbb{K}^d$,
\begin{align*}
\|T(\alpha)\|_X = \left\|\sum_{i=1}^{d} \alpha_i \, e_i\right\|_X \le \sum_{i=1}^{d} |\alpha_i| \, \|e_i\|_X \le C_1 \|\alpha\|_1,
\end{align*}
where $C_1 := \max_{1 \le i \le d} \|e_i\|_X$. Since $T$ is a continuous linear bijection between finite-dimensional normed spaces, and $\mathbb{K}^d$ is finite-dimensional, $T^{-1}$ is also continuous. (This follows from the fact that a continuous bijective linear map between finite-dimensional normed spaces has a continuous inverse, since the unit sphere in $\mathbb{K}^d$ is compact under $\|\cdot\|_1$ and $\|T(\alpha)\|_X$ attains a positive minimum $c > 0$ on $\{\alpha : \|\alpha\|_1 = 1\}$.) Therefore there exists $c > 0$ with
\begin{align*}
c \, \|\alpha\|_1 \le \|T(\alpha)\|_X \le C_1 \|\alpha\|_1 \quad \text{for all } \alpha \in \mathbb{K}^d.
\end{align*}
[guided]
To see that $T^{-1}$ is continuous, consider the function
\begin{align*}
f: S &\to \mathbb{R} \\
\alpha &\mapsto \|T(\alpha)\|_X,
\end{align*}
where $S := \{\alpha \in \mathbb{K}^d : \|\alpha\|_1 = 1\}$ is the unit sphere in $(\mathbb{K}^d, \|\cdot\|_1)$. The set $S$ is closed and bounded in $\mathbb{K}^d$, hence compact by the Heine-Borel theorem (since $\mathbb{K}^d$ is finite-dimensional). The map $f$ is continuous (as the composition of the continuous maps $T$ and $\|\cdot\|_X$). We claim $f > 0$ on $S$: if $f(\alpha) = \|T(\alpha)\|_X = 0$, then $T(\alpha) = 0$, hence $\alpha = 0$ by injectivity of $T$, contradicting $\|\alpha\|_1 = 1$. Since $f$ is a continuous positive function on a compact set, it attains its minimum:
\begin{align*}
c := \min_{\alpha \in S} \|T(\alpha)\|_X > 0.
\end{align*}
For arbitrary $\alpha \in \mathbb{K}^d \setminus \{0\}$, the vector $\alpha / \|\alpha\|_1 \in S$, so
\begin{align*}
\|T(\alpha)\|_X = \|\alpha\|_1 \cdot \left\|T\!\left(\frac{\alpha}{\|\alpha\|_1}\right)\right\|_X \ge c \, \|\alpha\|_1.
\end{align*}
The inequality also holds for $\alpha = 0$ (both sides are $0$). This gives $\|T^{-1}(y)\|_1 \le c^{-1} \|y\|_X$ for all $y \in Y$, confirming continuity of $T^{-1}$.
[/guided]
[/step]
[step:Show $Y$ is closed by showing limits of sequences in $Y$ remain in $Y$]
Let $(y_k)_{k \in \mathbb{N}}$ be a sequence in $Y$ converging to some $x \in X$, i.e., $\|y_k - x\|_X \to 0$. We must show $x \in Y$.
Write $y_k = T(\alpha_k)$ for a unique $\alpha_k \in \mathbb{K}^d$. Since $(y_k)$ is Cauchy in $(X, \|\cdot\|_X)$ and $T^{-1}$ is continuous with $\|\alpha_k - \alpha_j\|_1 \le c^{-1} \|y_k - y_j\|_X$, the sequence $(\alpha_k)$ is Cauchy in $(\mathbb{K}^d, \|\cdot\|_1)$. Since $\mathbb{K}^d$ is complete, $\alpha_k \to \alpha_*$ for some $\alpha_* \in \mathbb{K}^d$. By continuity of $T$,
\begin{align*}
y_k = T(\alpha_k) \to T(\alpha_*) \in Y.
\end{align*}
Since limits in a normed space are unique and $y_k \to x$, we conclude $x = T(\alpha_*) \in Y$. Therefore $Y$ is closed.
[guided]
Let $(y_k)_{k \in \mathbb{N}}$ be a sequence in $Y$ with $y_k \to x$ in $(X, \|\cdot\|_X)$. We need to show $x \in Y$.
Each $y_k$ has a unique coordinate representation $y_k = T(\alpha_k)$ with $\alpha_k \in \mathbb{K}^d$. Since $y_k \to x$, the sequence $(y_k)$ is Cauchy. Using the lower bound from the norm equivalence:
\begin{align*}
\|\alpha_k - \alpha_j\|_1 = \|T^{-1}(y_k) - T^{-1}(y_j)\|_1 = \|T^{-1}(y_k - y_j)\|_1 \le c^{-1} \|y_k - y_j\|_X.
\end{align*}
Since the right-hand side tends to $0$ as $k, j \to \infty$, the sequence $(\alpha_k)$ is Cauchy in $(\mathbb{K}^d, \|\cdot\|_1)$. The space $\mathbb{K}^d$ with any norm is complete (this is a standard fact for finite-dimensional normed spaces over $\mathbb{R}$ or $\mathbb{C}$), so $\alpha_k \to \alpha_*$ for some $\alpha_* \in \mathbb{K}^d$.
Applying the continuous map $T$:
\begin{align*}
y_k = T(\alpha_k) \to T(\alpha_*).
\end{align*}
But also $y_k \to x$ in $X$. Since limits are unique in a normed space, $x = T(\alpha_*) = \sum_{i=1}^{d} (\alpha_*)_i \, e_i \in Y$.
The essential mechanism: convergence in $X$ is transferred to convergence of the coordinate vectors via the continuous inverse $T^{-1}$, completeness of $\mathbb{K}^d$ provides a limit in the coordinate space, and $T$ maps this limit back into $Y$. This argument fails for infinite-dimensional subspaces because the coordinate-to-norm equivalence (with $c > 0$) relies on compactness of the unit sphere, which is available only in finite dimensions.
[/guided]
[/step]
