In a Banach space, can there be a closed subspace which does not have a closed complement?

As was pointed out by Kavi Rama Murthy, there is a theorem, already $50$ years old by now, proved by Lindenstrauss and Tzafriri, asserting that a Banach space $X$ is isomorphic to Hilbert space if and only if every closed subspace of $X$ is complemented. Therefore, every infinite dimensional Banach space not isomorphic to Hilbert space, contains uncomplemented subspaces. For a survey on the topic, see here: A survey on the complemented subspace problem.

My comment might be a little too brief, so I'll extend it a touch. You can look all of this up in Kalton and Albiac's Topics in Banach Space Theory (section 2.3) for the details, but in essence:

If $X$ is a separable Banach space then we can construct a continuous operator $Q: \ell_1 \rightarrow X$. This is done by restricting ourselves first to the unit ball (since it's easy to extend out to the full Banach space after that's been established) and defining $Q$ in the "obvious" way; i.e.

$$Q(\xi) = \sum_{j=1}^\infty \xi_n x_n $$

where $\{x_n\}_{n \in {\mathbb N}}$ is a countable dense subset of $X\cap B_X$ and $B_X$ is the unit ball in $X$, and $\xi = (\xi_1, \xi_2, \ldots )$ is an element of $\ell_1$. We can quickly show that $Q$ is linear and has norm one, and with a little more work (this is bulk of the proof) that $Q(B_{\ell_1})$ is dense in $B_X$.

Now, since $X$ is separable it is isometrically isomorphic to a quotient of $\ell_1$, and we can reach our conclusion: suppose $X$ is not isomorphic to $\ell_1$ and consider the closed subspace $\mathop{ker} Q$. Since $X = \ell_1 / \mathop{ker} Q$ this is isomorphic to a closed subspace of $\ell_1$, which requires $X$ to be isomorphic to $\ell_1$, which is a contradiction.