basic sequence and strictly singular operators

Suppose that $T: X\to Y$ is strictly singular. Prove

1) In every infinite dimensional subspace $Z$ of $X$, there exists a normalized basic sequence $(x_n)$ such that $||Tx_n|| <2^{-n}$.
2) For every $\epsilon$ >$0$ there exists an infinite dimensional subspace $Z$ of $X$ such that $T|_Z$ is compact and has norm less than $\epsilon$.

In part $1$, I got a normalized sequence $(x_n)$ such that the $||Tx_n|| <2^{-n}$, but it need not be basic. I know that if $(x_n)$ is weakly null, then it has a basic sub-sequence and I'm done. Can I choose it to be weakly null? If yes, why? if no, do you have any other idea?
Thank you .

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For part 1, take a normalised basic sequence $(z_n)$ in $Z$ and obtain the sequence $(x_n)$ as a block basic sequence of $(z_n)$. For part 2, consider the closed linear span of the block basic sequence obtained in part 1. –  Philip Brooker Mar 25 '13 at 12:50
I can't see why this block sequence satisfies the condition in part 1?? –  user61965 Mar 25 '13 at 14:40

Hint: Look of the proof that every Banach space contains a basic sequence. It uses Mazur's Lemma. Under the added assumption of $T$ being strictly singular (thus not bounded bellow on any infinite dimensional subspace), you can choose the vector $x$ in the Mazur's Lemma such that $||Tx||$ is as small as you wish. Use inductively Mazur's Lemma to obtain $(x_n)$ a basic sequence with the required property.
For 2) , choose appropriately the $x_n$ in 1) ($2^{-n}$ is not good enough) such that $SP_n$ converges to $S$, where $P_n$ are the coordinate projection for the basic sequence $(x_n)$, $Z$ is the closed span of $(x_n)$, and $S$ is the restriction of $T$ to $Z$. Since $SP_n$ is finite rank....