Continuous iff composition with every linear functional is continuous Let $X$ and $Y$ be normed spaces, $T:X \rightarrow Y$ a linear operator. Show that $T$ is continuous if $y' \circ T$ is continuous for every $y' \in Y'$. My idea is the following:
Suppose $T$ is not continuous, then there exists a sequence $x_n$ with $\|x_n\|=1$ such that $\|Tx_n\|>n$. If I could assure there is a subsequence $x_{n_k}$ such that $Tx_{n_k}$ are linearly independent I think I could construct an element $z'$ of $Y'$ such that $|(z' \circ T)x_{n_k}|>n_k$.
The first thing is, can I assure the existence of that subsequence? And if I can, how do I conclude applying H-B?
I'd like a solution with this same approach, but if there is a simpler proof i'd like to see it too.

I have a proof using the closed graph theorem:
Let $y' \circ T$ be continuous for all $y' \in Y'$. Define the operator $T^* : Y' \to X'$ by $y' \mapsto y' \circ T$. Obviously, $T^*$ is linear. Moreover, it is not hard to show that $T^*$ is closed. Hence, it is bounded by the closed graph theorem, i.e., $\|T^*\| \le C$. This implies
$$y'(T(x)) = (y'\circ T)(x) = (T^*y')(x) \le C \, \|y'\| \, \|x\|.$$
Hence, the norm of $T$ is bounded as well by $C$.
Note that $T^*$ is just the adjoint of $T$.
