This follows from the uniform boundedness principle. Consider the closed unit ball $B$ in $X$. Let $E=\{Tx:x\in X, \|x\|\leq1\}$. Then $y'(E)=(y' \circ T)(B)$ is bounded $\forall y' \in Y'$ by the continuity of $y' \circ T$. By the uniform boundedness principle, this implies that $E$ must be bounded, proving the continuity of $T$.
You are making a fair point, but there is no harm done: the proof is correct, but (in my humble opinion) the person that posted this has made a small mistake (a typo, actually): it should have been
$$a_j=\Lambda_j(Tw),\;\;\;w\in W$$
followed by
$|\Lambda_j(Tw)|\leq c\|Tw\|\leq c\|T\|\|w\|$, so that the functionals $\Lambda_j$ are defined on $T(W)$ which is finite dimensional (thus they are immediately continuous). The key is then to consider the functionals $\Lambda_j\circ T:W\to\mathbb{F}$ which are now immediately continuous as compositions of continuous maps and extend those on the entire domain.
I will add this as a comment to the answer in the original post too.
Edit: In the comments, OP asks what can we say about $\|T\|$ and $\|T'\|$ when $T$ is onto, where $T'$ denotes the extension. If $T$ is onto then we can assume without loss of generality that $Y=\mathbb{C}^n$ for some $n\geq1$, since $T$ is of finite rank. We equip $Y$ with the maximum norm, since all norms are equivalent. Let's look at the construction:
we begin with $T:W\to\mathbb{C}^n$ and we compose with the functional $\varepsilon_i:\mathbb{C}^n\to\mathbb{C}$ acting as $(z_1,\dots,z_n)\mapsto z_i$, so $\epsilon_i\circ T:W\to\mathbb{C}$ are bounded functionals. By Hahn-Banach we extend those to functionals $\varphi_i:X\to\mathbb{C}$ in a norm-preserving fashion, i.e. $\|\varphi_i\|=\|\varepsilon_i\circ T\|$. We now define $T':X\to\mathbb{C}^n$ as $T'(x)=(\varphi_1(x),\dots,\varphi_n(x))$. This is bounded, extends $T$ and observe that
$$\|T'\|=\sup_{x\neq0}\frac{\|T'(x)\|}{\|x\|}=\sup_{x\neq0}\sup_{1\leq j\leq n}\frac{|\varphi_j(x)|}{\|x\|}=\sup_j\sup_x\frac{|\varphi_j(x)|}{\|x\|}=\sup_{1\leq j\leq n}\|\varphi_j\|=\sup_{1\leq j\leq n}\|\varepsilon_j\circ T\|$$
But $\|\varepsilon_j\circ T\|\leq\|T\|$, so $\|T'\|\leq \|T\|$. On the other hand, $T'$ extends $T$, so obviously $\|T\|\leq\|T'\|$.
Best Answer
I'm going to revise the notation to make things easier to follow.
Say $X^*$ is the space of bounded linear functionals on $X$, and similarly for $Y^*$. I'm going to write $x$ and $y$ for elements of $X$ and $Y$ and I'm going to write $x^*$ and $y^*$ for elements of $X^*$ and $Y^*$.
Define the adjoint $T^*:Y^*\to X^*$ as usual: $$T^*y^*=y^*T.$$We can use the Closed Graph Theorem to show that $T^*$ is bounded (note that $X^*$ and $Y^*$ are Banach spaces). Assume that $y_n^*\to y^*$ and $T^*y_n^*\to x^*$; we need to show that $x^*=T^*y^*$. By definition that means we need to show that $x^*(x)=T^*y^*(x)$ for every $x\in X$. But since norm convergence in the dual implies pointwise convergence, $$x^*(x)=\lim T^*y_n^*(x)=\lim y_n^*(Tx)=y^*(Tx)=T^*y^*(x).$$
So $T^*$ is bounded. This implies that $T$ is bounded. NOTE that the Hahn-Banach theorem applies to incomplete normed vector spaces, in particular if $y\in Y$ then $$||y||=\sup_{||y^*||=1}|y^*(y)|.$$So for $x\in X$ we have $$||Tx||=\sup_{||y^*||=1}|y^*Tx|=\sup_{||y||=1}|T^*y^*x| \le \sup_{||y||=1}||T^*y^*||\,||x||\le||T^*||\,||x||.$$