What you are looking for is whether or not the following holds:
Let $X$ and $Y$ be normed spaces, either both real or both complex, such that $\dim X = \infty$ and such that $Y$ has non-zero vectors. Then there exists an unbounded linear operator $T \colon X \rightarrow Y$.
Let us attempt a proof.
Let $X$ be an infinite-dimensional real normed space.
Let $S$ be a linearly independent, countably infinite, ordered subset of $X$, say,
$$ S = \left\{ \, x_1, x_2, x_3, \ldots \, \right\}. $$
This $S$ can be extended to a basis (in fact an ordered basis) $B$ for $X$, using the choice axiom.
Now let us define a linear operator $T \colon X \rightarrow \mathbb{R}$ as follows:
$$ T\left( x_n \right) \colon= n \qquad T (x) \colon= 0 \ \mbox{ if } x \in B \setminus S. $$
Of course, a linear operator is uniquely determined by its values at the elements of a basis of its domain.
Now let us assume without any loss of generality that $$ \lVert x_n \rVert_X = 1 $$
for every $n \in \mathbb{N}$.
Then for any $n \in \mathbb{N}$, we have
$$ \lVert T \rVert \geq n. $$
Therefore
$$ \lVert T \rVert = \infty. $$
Now let us generalise this.
Let $X$ and $Y$ be normed spaces, either both real or both complex, such that $\dim X = \infty$ and such that $Y$ has elements other than the zero vector $\mathbf{0}_Y$.
Let $B$ be an ordered basis for $X$ such that $B$ contains a countably infinite subset
$$ S = \left\{ \, x_1, x_2, x_3, \ldots \right\} $$
such that
$$ \left\lVert x_n \right\rVert_X = 1 $$
for all $n \in \mathbb{N}$.
And, let $y_o$ be a non-zero vector in $Y$.
Let us now define $T \colon X \rightarrow Y$ as follows:
$$ T \left( x_n \right) \colon= n y_0 \qquad T(x) = \mathbf{0} \ \mbox{ if } x \in B \setminus S. $$
Then
$$ \lVert T \rVert = \infty. $$