Show that if a linear operator between Banach spaces is not onto, then its image is a meager set 
Let $X,Y$ be two Banach Spaces and let $T:X\to Y$ be a bounded linear operator. Show that either $T$ is onto or else $T(X)$ is a meager set.

I assumed that $T$ is not onto .Then we need to show that $T(X)$ is meager i.e. it is a set of first category. 
Though my attempt is useless,I am still giving it.Will someone please help me how to proceed here.
 A: To have a self-contained answer, I'll summarize the proof following the handout pointed out in a comment. Let $B_X$ and $B_Y$ be the open unit balls in $X$ and $Y$, respectively. 
Case 1: $\overline{T(B_X)}$ has empty interior. Then  it's a nowhere dense closed set, which allows us to cover $T(X)$ with countably many such sets: $T(X)\subset \bigcup_{n=1}^\infty \overline{T(nB_X)}$. Thus $T(X)$ is meager (of the 1st category).
Case 2: $\overline{T(B_X)}$ has nonempty interior. Since it's convex and symmetric about $0$, it follows that $\overline{T(B_X)}\supset rB_Y$ for some $r>0$. We will prove that $T(B_X)$ itself contains a neighborhood of $0$ in $Y$, which will imply $T$ is onto. 
Given $y\in B_Y$ with $\|y\|<r/2$, pick $x_1\in \frac12 B$ such that $\|y-Tx_1\|<r/4$. Then pick $x_2\in \frac14 B$ such that $\|y-Tx_1-Tx_2\|<r/8$, and so on. The resulting series $\sum x_n$ converges to $x\in B$ and $Tx=y$ by construction.
The last part of the proof is what we do in the proof of the Open Mapping Theorem; and indeed, in the lecture notes the result about $TX$ is meager is incorporated into a stronger form of  the OMT: if a bounded linear operator between Banach spaces has a nonmeager image, then it is an open map.
