Prove: $T \in L(Y,\ell^\infty(\mathbb K))$ is bounded $\implies$ $\forall n \in \mathbb N:g_n \in L(Y,\mathbb K)$ is bounded

1. ($T$ is bounded) $\rightarrow$ ($\forall n \in \mathbb{N}:g_n$ is bounded )

Assume $T$ is bounded, i.e. $$ \exists M \geq 0:\forall y\in Y: \|Ty\|_{\infty} \leq M\|y\|_{\infty}. $$ Since $\|Ty\|_\infty = \sup_{n\in\mathbb{N}}|g_n(y)| \leq M$, it follows $\exists M \geq 0:\forall y\in Y: \sup_{n\in\mathbb{N}}|g_n(y)| \leq M\|y\|_{\infty}$. But $\forall n \in \mathbb N: |g_n(y)| \le \sup_{n\in\mathbb{N}}|g_n(y)|$, that yields \begin{align} \exists M \geq 0:\forall n \in \mathbb N:\forall y\in Y: |g_n(y)| \leq M\|y\|_{\infty} \end{align} which is the same as saying each $g_n$ is bounded

Added:

2. ($\forall n \in \mathbb{N}:g_n$ is bounded ) $\rightarrow$ ($T$ is bounded)

Since for every $y$ in $Y$ we have $Ty \in \ell^\infty$, then $\forall y \in Y: \|Ty\|_\infty = \sup_{n\in \mathbb N} |g_n(y)| < \infty$. Also since $Y$ is Banach and $(g_n)$ are bounded we have by Banach-Steinhaus theorem (unifrom boundedness principle) that $\sup_{n \in \mathbb N}\|g_n\| < \infty$. So $$ \sup_{y:\|y\| =1} \|Ty\|_\infty = \sup_{y:\|y\| =1} (\sup_{n \in \mathbb N}|g_n(y)|) = \sup_{n \in \mathbb N} (\sup_{y:\|y\| =1}|g_n(y)|) = \sup_{n \in \mathbb N} \|g_n\| < \infty $$ (since both suprema exist,) so it follows $T$ is bounded.

(1) is true. Let $y\in Y$. Let $n\in\mathbb{N}$ be arbitrary, then $|g_{n}(y)|\leq||Ty||_{\infty}\leq||T||\cdot||y||$. This shows that $g_{n}$ is bounded and $||g_{n}||\leq||T||$.

(2) is true. For each $n\in\mathbb{N}$, define $T_{n}:Y\rightarrow l^{\infty}$ by $T_{n}y=(g_{1}(y),g_{2}(y),\ldots g_{n}(y),0,0,\ldots)$. Clearly $T_{n}$ is linear. Let $y\in Y$, then \begin{eqnarray*} ||T_{n}y|| & = & \max_{1\leq k\leq n}|g_{k}(y)|\\ & \leq & \max_{1\leq k\leq n}||g_{k}||\cdot||y||\\ & = & M||y|| \end{eqnarray*} where $M=\max_{1\leq k\leq n}||g_{k}||<\infty$. Therefore, $T_{n}$ is a bounded linear map. Consider the family of bounded linear maps $\{T_{n}\mid n\in\mathbb{N}\}$. We assert that for each $y\in Y$, $\{T_{n}y\mid n\in\mathbb{N}\}$ is a bounded subset of $l^{\infty}$. Let $y\in Y$. Since $T:Y\rightarrow l^{\infty}$ is a well-defined linear map, $Ty\in l^{\infty}$. That is, $M_{1}:=\sup_{k\in\mathbb{N}}|g_{k}(y)|<\infty$. For each $n$, we have that $||T_{n}y||_{\infty}=\sup_{k\leq n}|g_{k}(y)|\leq M_{1}$. This shows that $\{T_{n}y\mid n\in\mathbb{N}\}$ is a bounded subset of $l^{\infty}$. By Uniform Boundedness Principle, $M_{2}:=\sup_{n}||T_{n}||<\infty$. Let $y\in Y$ be arbitrary. For each $n\in\mathbb{N}$, we have that $\sup_{k\leq n}|g_{k}(y)|=||T_{n}y||_{\infty}\leq M_{2}||y||$. It follows that $||Ty||_{\infty}=\sup_{k}|g_{k}(y)|\leq M_{2}||y||$. Hence, $T$ is bounded.