Let $T:X\to Y$ be linear and continuous between Banach spaces $X,Y$. Suppose that $\overline{T(X)}=Y$ and $T(X)\neq Y$. Claim: there exists a $y\in Y$, such that for every $x_n$ with $Tx_n\to y$, $$\|x_n\|\to \infty,\quad n\to \infty.$$
Since $T(X)\neq Y$, $T$ is not surjective and therefore not an open mapping. I think I have to find a contradiction to this but not getting anywhere.
Here is a proof if $X$ is reflexive. Take any $y \in Y\setminus T(X)$ and $(x_n)_n \in X$ with $Tx_n \to y$. Suppose, for the sake of contradiction, that there was some $C > 0$ so that $||x_n|| \le C$ for infinitely many $n$. By passing to a subsequence, we may assume $||x_n|| \le C$ for each $n$. Recall that each $x_n \in X$ naturally corresponds to an element $\hat{x_n} \in X^{**}$ given by $\hat{x_n}(f) = f(x_n)$. By the Banach-Alaoglu theorem, since $||\hat{x_n}|| = ||x_n||$, there is some $\Phi \in X^{**}$ such that $\hat{x_n} \to \Phi$, i.e., for each $f \in X^*$, $\hat{x_n}(f) \to \Phi(f)$. Now if $X$ is reflexive, $\Phi = \hat{x}$ for some $x \in X$. Then, for any $g \in Y^*$, $$g(y) = \lim_n g(Tx_n) = \lim_n (g\circ T)(x_n) = \lim_n \hat{x_n}(g\circ T) = \hat{x}(g\circ T) = g(Tx),$$ so since linear functionals separate points, we see that $Tx = y$, a contradiction.
Lemma. Let $X,Y$ be Banach spaces and $T : X \to Y$ a bounded linear map. If $T(B_X(0,1))$ is not nowhere dense in $Y$, then $T$ is surjective.
Proof. Assume that $\overline{T(B_X(0,1))}$ has nonempty interior so it contains an open ball $B_Y(z,\delta)$ in $Y$. We then also have $B_Y(-z, \delta) \subseteq \overline{T(B_X(0,1))}$ so $B_Y(0,\delta) \subseteq \overline{T(B_X(0,1))}$.
We will show that $B_Y\left(0, \frac\delta2\right) \subseteq T(B_X(0,1))$. Let $y \in Y$, $\|y\| \le \frac\delta2$.
We have $2y \in B_Y(0, \delta) \subseteq \overline{T(B_X(0,1))}$ so there exists $x_1 \in B_X\left(0, \frac12\right)$ such that $\|2y-2Tx_1\| < \frac\delta2$, or $\|y-Tx_1\| < \frac\delta4$. Similarly there exists $x_2 \in B_X\left(0, \frac14\right)$ such that $\|y-Tx_1-Tx_2\| < \frac\delta8$. Continuing inductively we obtain a sequence $(x_n)_n$ in $X$ such that $\|x_n\| \le \frac1{2^n}$ and $\left\|y-\sum_{k=1}^n Tx_k\right\| < \frac\delta{2^{n+1}}$. Therefore $x = \sum_{n=1}^\infty x_n \in X$ and $Tx = y$.
Now since $T(B_X(0,1))$ contains an open ball, $T(X)$ has nonempty interior so $T(X) = Y$.
In our case $T$ is not surjective so $T(B_X(0,1))$ is nowhere dense in $Y$. The same holds for $T(B_X(0,m))$, $m\in\mathbb{N}$.
By passing to a subsequence we get
\begin{align} S &:= \{y \in Y : \exists(x_n)_n \text{ sequence in } X \text{ such that } Tx_n \to y \text{ and } \|x_n\| \not\to\infty\}\\ &\subseteq \{y \in Y : \exists(x_n)_n \text{ sequence in } X \text{ such that } Tx_n \to y \text{ and } (x_n)_n \text{ is bounded}\}\\ &\subseteq \bigcup_{m\in\mathbb{N}} \{y \in Y : \exists(x_n)_n \text{ sequence in } X \text{ such that } Tx_n \to y \text{ and } \|x_n\| \le m, \forall n\in\mathbb{N}\} \\ &\subseteq \bigcup_{m\in\mathbb{N}}\overline{T(B_X(0,m))} \end{align}
so $S$ is of first category in $Y$. Therefore $$\emptyset \ne S^c= \{y \in Y : \forall (x_n)_n \text{ sequence in } X \text{ such that } Tx_n \to y \text{ we have } \|x_n\| \to\infty\}$$
Moreover, $S^c$ is dense in $Y$.
Assume the statement does not hold. Then every $y\in Y$ is the limit of a sequence $(Tx_n)$ where $(x_n)$ is bounded. That is, $y$ is in the closure of the image of a open ball around the origin. This shows $$ Y = \cup_{R\in \mathbb N} \overline{ TB_R(0) } $$ Looking in the standard proof of the open-mapping theorem shows that $T$ is open, which implies $T$ surjective, which is a contradiction.
(This is in essence mechanodroid's answer in reverse steps)
This only gives the existence of one $y$ for which $Tx_n\to y$ implies $\|x_n\|\to \infty$. The proof by mathworker21 shows this statement for all $y\in Y\setminus R(T)$.
In general such a statement is not true. Take $X=Y=c_0$, $$ T(x) = (x_1, x_2/2, x_3/3, \dots). $$ Then $T$ has dense range, and $T^{-1}$ is unbounded on the range (to see this consider $T^{-1}e_n$). Take $y=(1,1/2,1/3,\dots)$, which is not in the range of $T$. Define $x_n=(1,\dots, 1,0,\dots)$. Then $Tx_n\to y$ and $(x_n)$ is bounded.
