Folland, "Real Analysis", Chapter 5.3, Exercise 36.

The following version of the open mapping theorem holds: If $T:Y\to X$ is a bounded operator between Banach spaces with $B_X \subseteq c \,\overline{T(B_Y)}$ for some constant $c>0$ (and the unit balls $B_X$ and $B_Y$) then $T$ is surjective (and open). Sometimes such a $T$ is called almost open. (The proof is in every book on functional analysis, the first step in the proof of the classical formulation of the OMT is to show that $T$ is almost open.)

In your case, $T(B_Y)$ contains the dense subset $\lbrace x_n:n\in \mathbb N\rbrace$ of $B_X$.

Since $T$ is linear, it's enough to show that every $x\in X$ with $||x||\leq 1$ is in the image of $T$. To do this, we proceed inductively: there exists $x_{n_1}$ such that $||x-x_{n_1}||<\frac{1}{2}$. If $y=2(x-x_{n_1})$ then $||y||<1$, so there exists $x_{n_2}\neq x_{n_1}$ such that $||y-x_{n_2}||<\frac{1}{2}$, hence $||x-x_{n_1}-\frac{1}{2}x_{n_2}||<\frac{1}{4}$. In general, if $x_{n_1},\dots,x_{n_k}$ have been chosen such that $$ \Big|\Big|x-\sum_{j=1}^k2^{1-j}x_{n_j}\Big|\Big|<2^{-k}$$ then $y=2^k(x-\sum_{j=1}^k2^{1-j}x_{n_j})$ is in the unit ball, hence there exists $x_{n_{k+1}}\not\in\{x_{n_1},\dots,x_{n_k}\}$ such that $||y-x_{n_{k+1}}||<\frac{1}{2}$, i.e. $$ \Big|\Big|x-\sum_{j=1}^{k+1}2^{1-j}x_{n_j}\Big|\Big|<2^{-k-1}$$ Finally, define $f\in \ell^1(\mathbb{N})$ by $f(n)=2^{1-k}$ if $n=n_k$, $f(n)=0$ otherwise. Then $Tf=x$.