Okay I have been working under the assumption that this is "obvious" for a while now, but it started to bug me and now I'm fumbling to prove it.
Suppose $X$ is a normed linear space (possibly infinite dimensional). Let $\mathcal{B}$ be a Hamel basis of $X$. Fix $b \in \mathcal{B}$. Is the coordinate projection $P_b : X \to \mathbb{C}$ or $\mathbb{R}$ defined by $P_b(b) = 1$ and $P_b(x)=0$ for $x \notin \mathrm{span}(\{b\})$ continuous?
No. With notation as in the problem, suppose $X$ is infinite-dimensional and complete, and let $b_1, b_2, ...$ be a countable subset of $B$. WLOG $\| b_i \| = 1$. Then the series
$$\sum_i \frac{b_i}{2^i}$$
converges to some element of $X$ which, by assumption, can be expressed as a finite sum $\sum_j c_j b_j$, where $b_j \in B$ and the sum ranges over some finite index set $J$. It follows that
$$\sum_{i=1}^n \frac{b_i}{2^i} - \sum_j c_j b_j \to 0$$
as $n \to \infty$. From here it follows that all but finitely many (at most $|J|$) of the projections $P_{b_i}$ fail to be continuous because they do not preserve the above limit.
You should at least require that $B$ is a Schauder basis, but even then I think the conclusion fails.
Qiaochu Yuan's answer is neat but hinges on completeness and is not in the strongest form. Below I show that every normed space (not necessarily complete) has a Hamel basis such that every projection is discontinuous.
In what follows $X$ is a normed infinite dimensional space over reals.
Theorem. There exists a Hamel basis $\mathcal B$ of $X$ such that every coordinate projection is not continous.
Lemma. There is a linearly independent dense subset of $X$.
Sketch of a proof. Note that $\dim X \geq \text{dens}(X) = \text{weight}(X)$ and use transfinite induction to construct a required subset. Details are in this MSE answer by Nate Eldredge.
Proof of the theorem. By the above lemma there is a linearly independent dense subset $A\subset X$. Extend this subset to a Hamel base $\mathcal B$. Pick any $b \in \mathcal{B}$. Because $A \setminus \{b\}$ is still dense in $X$ (as it is a $T_1$ space without isolated points) we can find a sequence $(a_n)$ in $A \setminus \{b\} \subset \mathcal{B}$ such that $$ a_n \to b. $$ But $P_b$ does not respect this convergence $$ P_b(a_n) = 0 \to 0 \neq 1 = P_b(b). $$ Hence $P_b$ is discontinuous, which completes the proof.
Remark. If $X$ is a Banach space with a Schauder basis $(s_n)$ then every coordinate projection $P_{s_n}$ is continuous (details on page 32 of "Sequences and series in Banach spaces" by Diestel). But for a general normed space the analogous statement is not true. Bill Jonson gave a nice example on Mathoverflow: Take $(\ell_1, ||\cdot||_2)$. The sequence $(x_n)$ where $x_1 :=e_1$ and $x_n := e_1 + e_n$ for $n>1$ is a basis (here $e_k = (0,\dots, 0,1,0, \dots)$ and the only non-zero term is on k-th place). But $P_{x_1} = (1, -1,-1, \dots)$ is not $||\cdot||_2$-continous.
Remark. If $X$ is a finite-dimensional (Hausdorff) topological vector space and $Y$ is an arbitrary tvs, then all linear maps $T: X \to Y$ are continous. In particular, projections from finite dimensional normed spaces are continuous.
