Let $A \in L(X,Y)$, $\tilde y \in Y^*$, $x \in X$. Then from $\left(A^*(\tilde y)\right)(x) := \tilde y (A(x))$ you get
$$|A^*(\tilde y)(x)|= |\tilde y (A(x))|≤\|\tilde y\|_{Y^*} \|A(x)\|_Y ≤ \|\tilde y\|_{Y^*} \|A\|_{L(X,Y)} \|x\|_X$$
For $A^*$ to be bounded you must consider the following expression
$$\|A^*\|_{L(Y^*,X^*)}=\sup_{\tilde y \in Y^*, \ \|\tilde y\|_{Y^*}≤1} \left\{\|A^*(\tilde y)\|_{X^*}\right\}$$
Note that you have the following inequality, which follows from the first one
$$\|A^*(\tilde y)\|_{X^*}=\sup_{x \in X, \ \|x\|_X≤1}\{|A^*(\tilde y)(x)|\}≤\sup_{x \in X, \ \|x\|_X≤1}\{\|\tilde y\|_{Y^*} \|A\|_{L(X,Y)}\ \|x\|_X \}$$
Putting it together gives you
$$\|A^*\|_{L(Y^*,X^*)}≤\|A\|_{L(X,Y)}$$
Note further that the construction gives you $\|A^{**}\|_{L(X^{**},Y^{**})}≤\|A^*\|_{L(Y^*,X^*)}≤\|A\|_{L(X,Y)}$. If you restrict $A^{**}$ to the subspace given by the embedding of $X$ into $X^{**}$ you get precisely $A$ again (composed with the embedding of $Y$ into $Y^{**}$). So $\|A^{**}\|_{L(X^{**},Y^{**})}≥\|A\|_{L(X,Y)}$ also holds, and the two inequalities are equalities, so also $\|A^*\|=\|A\|$.
Your suspicion is correct: the topology on $(X\times Y,\lVert \cdot\rVert)$ may be strictly coarser than the product topology. Namely, consider a Banach space $(W,\lVert \cdot\rVert)$, a proper dense subspace $X$ and a subspace $Y$ such that $X\oplus Y=W$. Then consider $(X\times Y,\lVert \cdot\rVert)$ with the norm of $W$ (which is effectively $W$ under the identification $(x,y)=x+y$) versus the product space $(X,\left.\lVert \cdot\rVert\right\rvert_X)\times(Y,\left.\lVert \cdot\rVert\right\rvert_Y)$. $\lVert(x,0)\rVert=\lVert x\rVert_X$ and $\lVert(0,y)\rVert=\lVert y\rVert_Y$, but the topology is not the product topology because $X\times\{0\}$ is closed in the product, but not in $W$. In fact, since the product space has a non-Banach norm, it can't be homeomorphic to a Banach space at all.
More generally, with the same assumptions as your question plus $\lVert\cdot\rVert$ Banach, the topologies are the same if and only if both $X$ and $Y$ are Banach.
Added: It should be noted that $W$ being Banach is superfluous for the purpose of making a counterexample. The same can be done with any normed space that has a non-closed vector subspace. As a side remark, in this generality the product space may still be homeomorphic to $W$: see here an example which exploits the fact that pre-Hilbert spaces of algebraic dimension $\aleph_0$ are linearly isometric.
Best Answer
Let $x\in S_1$. Then $\|Tx\|\leq\|T\|\|x\|\leq\|T\|\cdot1=\|T\|$. So $T(S_1)\subset\{y\in Y: \|y\|\leq\|T\|\}$, so $T(S_1)$ is bounded.