Range of bounded operator is of first category

We can write $R(T)=\bigcup_{n=1}^{+\infty}T(B(0,n))$. We have to show that $T(B(0,n))$ is nowhere dense. Otherwise, the closure of $T(B(0,1))$ would have a non-empty interior, and we would have $B(0,r)\subset \overline{T(B(0,1)}$ for some $r>0$. Considering $T/r$ if necessary, we can assume that $r=1$. Let $y\in Y$ such that $y\notin R(T)$ of norm $1$. We can construct, by induction on $n$, a sequence $\{x_k\}\subset B(0,1)$ such that for each $n$, $$\left\lVert y-T\left(\sum_{j=1}^n\frac{x_j}{2^j}\right)\right\rVert\in (0,2^{-(n+1)}).$$ Indeed, take $x_1\in B(0,1)$ such that $\lVert y-Tx_1\rVert\leq \frac 12$. Then $y-Tx_1\notin R(T)$, and is of norm $\leq 1/2$. Take $x_2'$ of norm $\leq 1/2$ such that $Tx'_2$ approaches $y-Tx_1$ up to $1/4$ and take $x_2=2x_2'$. If $x_1,\dots,x_n$ are construct, we have that $y-T\left(\sum_{j=1}^n\frac{x_j}{2^j}\right)\notin R(T)$. Let $x'_{n+1}$ of norm $\leq 2^{-(n+1)}$ such that $$\left\lVert y-T\left(\sum_{j=1}^n\frac{x_j}{2^j}\right)-T'x_{n+1}\right\rVert\leq 2^{-(n+1)},$$ and take $x_{n+1}=2^{n+1}x'_{n+1}$. Let $s_n:=\sum_{k=1}^n2^{-k}x_k$. Since $\lVert x_k\rVert\leq 1$, we have $\lVert s_{m+n}-s_n\rVert\leqslant 2^{-n}$, hence $(s_n,n\geqslant 1)$ is a Cauchy sequence, which converges, by completeness of $X$ to some $x$. By boundedness of $T$, we would have $y=Tx$ (because $s_n\to x$ and $Ts_n\to y$), a contradiction.

I like Davide's answer a lot. But here is another answer that is essentially the same but is phrased slightly differently.

We prove the contrapositive of the desired assertion. That is, we prove that if $T(X)$ (=$R(T)$) is not of the first category in $Y$, then $T$ is surjective.

First observe that the proof of the usual open mapping theorem still works if instead of $Y$ Banach and $T$ surjective we assume that $Y$ is a normed space and $T(X)$ (=$R(T)$) is not of the first category in $Y$.

Thus if $T(X)$ (=$R(T)$) is not of the first category in $Y$, then $T$ is an open mapping from $X$ to $Y$. In particular, $T(X)$ is open in $Y$.

Now deduce that $T$ is surjective by using the following statement (whose proof is an easy exercise): If $V$ is a subspace of a normed space $W$, and $V$ contains an open subset of $W$, then $V=W$.