Bertini's Theorem small print Suppose $S\subset \mathbb{P}^n$ is a smooth del Pezzo surface and $C$ is an irreducible smooth curve (you can make it rational if it simplifies the setting) such that $\mathcal{L}=\vert -K_S-C\vert $ is non-empty. To simplify you may consider $-K_S$ is very ample and let me deal with the degree $1$ and $2$ cases. Moreover suppose that $h^0(S,\mathcal{O}(\mathcal{L}))\geq 2$ (i.e. $\mathcal{L}$ is at least a pencil).
It was my understanding that by Bertini's theorem one could choose a general member $L\in\mathcal{L}$ such that $L$ is smooth (and reduced and connected). I have been told this is wrong and after going to Hartshorne (and Wikipedia and some expository paper by Kleiman that Francesco added to the comments) I am also of the opinion that it may actually be wrong, but that $L$ must be irreducible away of the base locus of $\mathcal{L}$.
However I am unable of providing a proof nor a counter-example. Does someone have an insight on this? I also suspect the base locus of $\mathcal{L}$ may actually be empty.
Edit: Originally $H$ was a hyperplane section. The question is actually motivated by 'the' hyperplane section so I have rephrased it to meet this point. Apologies for the confusion.
 A: The point here is the following result, that you can find in Zariski's book "Algebraic Surfaces", page 26. Zariski calls it "Extended Theorem of Bertini".

Theorem (Extended Bertini)
(1) The general curve of an irreducible linear system cannot have multiple points outside the base locus of the system.
(2) A reducible linear system, without fixed components, is necessarily composed with the curves of a pencil.

Here "reducible" [resp. "irreducible"] means that the general curve of the system is reducible [resp. irreducible].
Now, let us write $\mathcal{L}=Z+\mathcal{M}$, where $Z$ is the fixed part and $\mathcal{M}$ is the moving part. Then by Extended Bertini it follows that the general element $M \in \mathcal{M}$ is necessarily irreducible, unless $\mathcal{M}$ is composed with a pencil.
The last situation can occur. For instance let $S=\mathbb{P}^1 \times \mathbb{P}^1$, whose natural pencils are denoted by $|F_1|$ and $|F_2|$, and take $H=F_1+2F_2$ and $C \in |F_1|$. Then $H$ is very ample but $\mathcal{L}=|H-C|=|2F_2|$, which is without fixed part and composed with the pencil $|F_2|$. In fact, any element of $|2F_2|$ is the union of two curves in the pencil $|F_2|$, in particular it is not irreducible.
Remark. The situation described in J. C. Ottem's comment is slightly different. In that example, indeed, we have a fixed part $Z=2E$; the moving part, however, is irreducible.
A: I think you are right  in the special case of a Del Pezzo surface $S$. Here's an idea of proof.
a)  we may assume that $K^2_S>1$. 
Proof:  since $K_S$ is ample, if $K^2_S=1$ then every curve of $|K_S|$ is irreducible.
b) if $|F|$ is an irreducible  pencil, then $F^2=0$, $FK_S=-2$, i.e. the general $F$ is a smooth projective curve. 
Proof:  by Riemann-Roch, we have  $2=h^0(F)=1+ (F^2-FK_S)/2$, namely $F^2-FK_S=2$. Since $F^2\ge 0$ and $-FK_S>0$, there are two possibilities:
$F^2=0, K_SF=-2$ or $F^2=1$ and $FK_S=-1$. The second possibility contradicts the index theorem, since $K^2_S>1$ by a).
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Now assume for contradiction that $-K_S=C+rF+Z$, where $C$ is smooth irreducible, $r>1$ is an integer, $|F|$ is an irreducible pencil and $Z$ is an effective divisor such that $|K_S-C|=Z+r|F|$. By b) we have $F^2=0$, $K_SF=-2$.
c) $S$ is not ${\mathbb P}^1\times {\mathbb P}^1$ or ${\mathbb P}^2$.
Proof: ${\mathbb P}^2$ has no free pencil $|F|$;  if $S={\mathbb P}^1\times {\mathbb P}^1$ the only possible $|F|$ are the two rulings of the product. 
d) $K^2_S\ge 5$
Proof: We have $K^2_S\ge -K_S(C+rF)\ge -K_SC+4\ge 5$, since $-K_S$ is ample.
e) $r\le 3$ and $K^2_S\ge 6$.
Proof: Assume that $S$ is the blow up of ${\mathbb P}^2$ at points $P_1,\dots P_k$, denote by $e_1,\dots e_k$ the corresponding exceptional curves and write $-K_S=3L-(e_1+\dots e_k)$, where $L$ is the class of a line in ${\mathbb P}^2$. Notice that $k\le 4$ by d).
The image of $|F|$ in ${\mathbb P}^2$ is either the  pencil of lines through, say, $P_1$ or the pencil of conics   through, say $P_1, \dots P_4$. Since $-K_S-2F$ is effective, the second case cannot occur. If $k=4$, using a Cremona transformation the two cases can be switched, so  $k=4$ (i.e., $K^2_S=5$) cannot occur either. In addition we have $r\le 3$, since $-K_S-rF$ is effective.
f) End of proof: the proof can be completed by enumeration, since the only possibility is that $S$ is the blow up of ${\mathbb P}^2$ at 1, 2 or 3 points and $|F|$ is the pencil of lines through one of the points.
