Suppose that $T$ is not continuous under norm convergence. Then there is a sequence $\{x_n\}$ such that $ \Vert x_n\Vert\le 1,\quad \text { and }\quad\Vert T(x_n)\Vert \ge n^2 \quad\text{ for each }n. $ (there is a sequence $z_n$ of non-zero vectors in $X$ converging in norm to $0$ for which $\lim\limits_{n\rightarrow\infty} \Vert Tz_n\Vert\ne 0$. Then $\{z_n/\Vert z_n\Vert\}$ is in the unit ball of $X$ and $\{T( z_n/\Vert z_n)\Vert\}$ is unbounded).
Now $\Vert x_n/n \Vert\rightarrow 0$, and thus $y^* (x_n/n)\rightarrow0$ for each $y^*\in Y^*$. Since $T$ is assumed to be continuous under weak convergence, $y^*T(x_n/n)\rightarrow 0\quad, \text{ for any }y^*\in Y^*. $
But then, it follows from the Uniform Boundedness Principle$^\dagger$ that $\{T(x_n/n)\}$ is norm bounded.
However, $\Vert T(x_n/n)\Vert\ge n$ for each $n$.
$^\dagger$ We consider
$\{Tx_n :n=1,2,\ldots\}$ as a subset of
$Y^{**}$. That is we consider
$Tx_n$ as a continuous linear functional on
$Y^*$ (which is a Banach space). The
$Tx_n$ are pointwise bounded, and, thus, norm bounded in
$Y^{**}$. But the norms of the
$Tx_n$ in
$Y^{**}$ are the same as the norms in
$Y$.